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1 Kinematics_______________________________________________________________________________
K I N E M A T I C S
SYLLABUS:
Motion in one and two dimension, Projectile motion, Relative Velocity & Circular Motion.
POSITION VECTOR
If the coordinates of a particle are given by (x2, y2, z2) its position vector with respect to (x1,y1,z1) is given
by →r = (x2 - x1) i + (y2 - y1) j +(z2 - z1) k . Usually, position vector with respect to the origin (0,0,0) is
specified and is given by →r =x i +y j +z k
DISPLACEMENT
Displacement is a vector quantity. It is the shortest distance between
the final and initial positions of a particle. If 1
→r is the initial
position vector and 2
→r is the final position vector, the displacement
vector is given by 12
→→→−=∆ rrr .
The magnitude of the displacement is given by
( ) ( ) ( )2122
122
12 zzyyxx −+−+−
This is nothing but the straight line distance between two points (x1,y1,z1) and (x2,y2,z2). The displacement is
independent of the path taken by the particle in moving from (x1 , y1, z1) to (x2, y2, z2)
DISTANCE :
If a particle moves along a curve, the actual length of the
path is the distance. Distance is always more than or equal
to displacement.
Illustration – 1 :
A car travels along a circular path of radius (50 /π)m with a speed of 10 m/s. Find its displacement and distance
after 17.5 sec.
Solution :
)z,y,x(P 111)z,y,x(Q 222
1
→
r
→
∆ r
2
→
r
2 Kinematics_______________________________________________________________________________
Distance = (speed) time = 10 (17.5) = 175 m
Perimeter of the circular path = 2 π (50/π) = 100 m
⇒ The car covers 14
3 rounds of the path
If the car starts from A, it reaches B and the displacement is the shortest
distance between A and B
⇒ Displacement = 22 RR + = π
=250
2 R m.
INSTANTANEOUS AND AVERAGE VELOCITY :
If ∆→r is the displacement of the particle in time ∆t, the average velocity is given by
→V average =
t
r
tt
rr
∆
∆=
−
−→→→
12
12
∆ = Find value - Initial value. The above definition is valid for any magnitude of ∆ large or small. But
when ∆ is infinitesimally small, the instantaneous velocity is obtained.
→V instantaneous = Lt
t 0→∆ t
r
∆
∆→
= dt
rd→
In normal notation, velocity refers to the instantaneous velocity.
SPEED :
Speed = time
cetanDis
When the time under consideration is very small, distance becomes equal to the displacement and speed
becomes the magnitude of instantaneous velocity. speed is represented only by its magnitude where as
velocity is represented by magnitude as well as direction.
INSTANTANEOUS AND AVERAGE ACCELERATION :
If →
∆ V is the change in velocity in time ∆t, average acceleration is given by
averagea→
= t
V
∆
∆→
.
A
B
3 Kinematics_______________________________________________________________________________
When ∆ becomes infinitesimally small, t
VLtt ∆
∆→
→∆ 0 = dt
Vd→
which gives the instantaneous acceleration.
In normal notation, acceleration refers to the instantaneous acceleration.
==
→→→
dt
rd
dt
d
dt
Vda =
2
2
dt
rd→
It may be noted here that magnitude of 2
2
dt
rd→
is not equal to 2
2
dt
rd always (as in the case of circular motion)
Illustration – 2 :
A bus shuttles between two places connected by a straight road with uniform speed of 36 kmph. If it stops at each
place for 15 minutes and the distance between the two places is 60 km, find the average values of
(a) Speed (b) Velocity
(c) acceleration between t = 0 and t = 2 hours and the instantaneous values of
(d) Velocity (e) acceleration at t = 2 hrs.
Solution :
Time taken for forward trip = 36
60 =
3
5 hrs.
Time of stoppage = 15 min = 0.25 hrs.
Time available for return trip = 2 - 5/3 - 0.25 = 1/12 hrs.
Distance travelled in the return trip = (36) 1/12 = 3 km.
a) Average speed = timeTotal
distanceTotal =
2
360 + = 31.5 kmph.
b) Average velocity = Time
ntDisplaceme =
2
360 − = 28.5 kmph
c) Average acceleration = Time
velocityinchange
kmd 31 =
kmd 60=
4 Kinematics_______________________________________________________________________________
= t
VV→→
− 12 = ( ) ( )
2
3636 +−− = - 36 km/H2 = -
360
1 m/s2
d) Velocity at t = 2 hours = - 36 kmph
e) Acceleration at t = 2hours = 0 as there is no change in velocity
Illustration – 3 :
A car travels towards North for 10 minutes with a velocity of 60 Kmph, turns towards East and travels for 15
minutes with a velocity of 80 kmph and then turns towards North East and travels for 5 minutes with a velocity of
60 kmph. For the total trip, find a) distance travelled b) displacement c) average speed d) average velocity
and e) average acceleration.
Solution :
Total time taken = (10 + 15 + 5)min = 1/2 hour
a) Distance travelled = d1 + d2 + d3
= 60
60
10 + 80
60
15 + 60
60
5
= 10 + 20 + 5 = 35 km
b) displacement →→→→
++= 321 SSSS
= 10 j + 20 i +5 cos 450 i + 5 sin 45 j
= 23.5 i + 13.5 j
Magnitude of displacement = ( ) ( )22 513523 .. + ~ 27 km
c) Average speed =
hr
kmdiTotal
=
2
1
35
Time
travelled stance = 70 kmph.
d) Average velocity = Time
ntDisplaceme = ( )ji
j.i.2747
2
1
513523+=
+ kmph.
at an angle θ with the East given by Tan θ = 523
513
.
.
Magnitude of average velocity = 22 2747 + = 54 kmph
e) Average acceleration = Time
yin velocit change
1
→
S2
→
S→
S
3
→
S
5 Kinematics_______________________________________________________________________________
= Time
velocityInitialvelocityFinal −
= ( ) ( )
2
6045604560 0 jjsinicos −+
= ( )ji 921 − km/H2
at an angle θ with the East given by Tanθ = 21
9−
Magnitude of average acceleration = 22 921 + ~ 23 km/H2 ~ 1.8 x 10-3 m/s2
Illustration – 4 :
A car moving along a circular path of radius R with uniform speed covers an angle θ during a given time. Find its
average velocity and average acceleration during this time.
Solution :
Let V be the speed of the car
V = time
Distance =
t
Rθ where θ is in radians.
Displacement= θ−+ cosRRR 222 2 from the triangle OAB
= 2R sin θ/2
Average velocity = Time
Diplacemnt =
θ
θ
V
R
sinR2
2 =
θ
θ
22 sinV
Average acceleration = t
V
time
velocityinChange ∆=
∆V = θ−+ cosVVV 222 2 = 2 V sin 2
θ
⇒ Average acceleration =
θ
θ
V
R
sinV2
2 =
θ
θ
R
sinV2
2 2
When θ is small sinθ ~ θ and
Average velocity =θ
θ
22 sinV
= V2
V2
=θ
θ
A V
B
Vo
θ
V
V
θV∆
6 Kinematics_______________________________________________________________________________
⇒ Average velocity = Instantaneous velocity for small angular displacements
Average acceleration = θ
θ
=θ
θ
R
V
R
sinV2
22
222
= R
V2
⇒ Average acceleration = Instantaneous acceleration for small angular displacements.
KINEMATICAL EQUATIONS : ( CONSTANT ACCELERATION ) :
V = u + at
V2 - u2 = 2aS
S= ut + 1/2 at2
The above equations are valid only for constant acceleration and in a particular direction. u,v and s must be
taken with proper sign. Usually the direction of u is taken as positive and the sign of other variables are
decided with respect to this direction.
Displacement during the nth second = Sn - Sn-1
= u + 2
a (2n - 1)
It may be noted here that this is not the distance travelled in the nth second.
Illustration – 5 :
A particle is vertically projected upwards with an initial velocity of 22.5 m/s. Taking g=10 m/s2 find a) velocity
b) displacement c) distance travelled in t = 4 sec and d) displacement and distance travelled in 3rd second
Solution :
Taking the upward direction positive
a) V = u + (-g) t = 22.5 - 10 (4) = -17.5 m/s ⇒ 17.5 m/s down wards
b) S = ut + 1/2 (-g) t2 = 22.5 (4) - 1/2 (10) 42 = 10 m
c) Time to reach the top most point = t0 and at the top most point velocity
becomes zero.
V = u - gt0 ⇒ 0 = 22.5 - 10 (t0) ⇒ t0 = 2.25 sec
Distance travelled in 4 sec = d1 + d2
d1 = u t0 - 1/2 g 20t = 22.5 (2.25) - 1/2 (10) (2.25)2 = 25.3 m
d1 can be found from V2 - u2 = 2a S also.
0 - (22.5)2 = 2(-10) d1 ⇒ d1 = ( )
20
522 2. = 25.3 m
1d
2d
7 Kinematics_______________________________________________________________________________
d2 can be found from S = ut + 1/2 at2 applied along the down ward direction starting from the top
most point
d2 = 0 (t - t0) + 1/2 g (t - t0)2 = 1/2 (10) (4 - 2.25)2 = 15.3 m
⇒ Distance travelled in 4 sec = 25.3 + 15.3 = 40.6 m
Displacement in 4 sec = d1 - d2 = 25.3 - 15.3 = 10 m
Displacement can also be found directly by applying S = ut + 1/2 at2 along the vertical
Displacement in 4 sec = 22.5 (4) - 1/2 (10) (4)2 = 10M
d) 3rd second is from t = 2 sec to t = 3 sec.
Displacement in the 3rd second = u + 2
a (2n - 1)
= 22.5 - 2
10 (6 - 1) = -2.5 m
When there is no change in the direction of the motion along a straight line, distance will be equal to
displacement. When the particle reverses its direction during the time under consideration, distance will be
more than the displacement and the time at which the reversal is taking place must be found.
When the particle reverses its direction, its velocity becomes zero.
using V = u + at, 0 = 22.5 - 10 (t0) ⇒ t0 = 2.25 sec
d = d1 + d2
using the formula S = ut + 1/2 at2
d1 = [22.5 (2.25) - 1/2 (10) (2.25)2 ] - [22.5 (2) - 1/2 (10) (2)2]
= 0.31 m
Along the downwards vertical starting from the top
d2 = 0 (3 - 2.25) + 1/2 (10) (3 - 2.25)2 = 2.81 m
⇒ d = 0.31 + 2.81 = 3 .12 m
KINEMATICAL EQUATIONS ( VARIABLE ACCELERATION ) :
When the acceleration is variable, the kinematical equation take the form
V = dt
dx a =
2
2
dt
xd
dt
dV=
a = dx
VdV
dt
dx
dx
dV=
V = u + ∫
t
dta
0
sect 3=sect 2=
sec.t 252=
1d2d
8 Kinematics_______________________________________________________________________________
x = ut + dtdta
tt
∫∫00
and V2 - u2 = 2 ∫x
dxa
0
Illustration – 6 :
The position coordinate of a particle moving along a straight line is given by x = 4 t3-3t2+4t+5. Find a) Velocity
and acceleration as a function of time b) Displacement as a function of time c) the time at which velocity becomes
zero and the acceleration at this time d) the time at which acceleration becomes zero and the velocity at this time.
Solution :
a) V = dt
dS =
( )dt
xxd 0− =
dt
dx Where x0 is the initial position coordinate which is a constant
= dt
d (4t3 - 3t2 + 4t + 5) = 12 t2 - 6t + 4
a = dt
dv =
dt
d (12t 2 - 6t + 4) = 24t - 6
b) Displacement = (position coordinate at time t) - (position coordinate at t = 0)
= (4t3 - 3t2 + 4t + 5) - (5)
= 4t3 - 3t2 + 4t
c) When V = 0, 12 t2 - 6t + 4 = 0 ⇒ t = 12
4893 −±
since this value is imaginary, the velocity never becomes zero.
d) When a = 0, 24t – 6 =0 and t = 24
6 =
4
1 units and the velocity of the particle at this time, V=
12 4
134
4
16
4
12
=+
−
units
Illustration – 7 :
The velocity of a particle moving in the positive direction of the x axis varies as V = α x where α is a positive
constant. Assuming that at the moment t = 0 the particle was located at the point x = 0, find a) the time
dependence of the velocity and the acceleration of the particle b) the mean velocity of the particle averaged over
the time that the particle takes to cover the first S meters of the path.
Solution :
9 Kinematics_______________________________________________________________________________
a) V = dt
dx = α x ⇒ ∫∫ α=
tx
dtx
dx
00
2 x = α t and x = 4
22 tα
⇒ V = 2
2 tα and a =
dt
dV =
2
2α
b) Mean velocity = time
ntDisplaceme
Displacement = S, and the time taken for this displacement t = α
S2
Mean velocity = ( )
α
S
S
2 =
2
Sα
Mean velocity can also be found from the following formulae
V mean =
∫∫
dx
dxV when V is a function of x
and V mean =
∫∫
dt
dtV when V is a function of time
KINEMATICAL EQUATIONS IN VECTOR FORM ( CONSTANT ACCELERATION ) :
tauV→→→
+=
→→→→→→
=− S.au.uV.V 2
2
2
1tatuS
→→→+=
The above equations are useful in 2 and 3 dimensional motion.
Illustration – 8 :
A particle moving on a horizontal plane has velocity and acceleration as shown in the diagram at time t = 0. Find
the velocity and displacement at time 't'.
10 Kinematics_______________________________________________________________________________
Solution :
METHOD - I
→u = u cos300 i + u cos 60 j =
2
3 u i + j
u
2
→a = - a cos 45 i - a cos45 j =
2
a− i -
2
a j
→V = tau
→→+ = it
au
−
22
3 + jt
au
−
22
The magnitude of the velocity =
22
2222
3
−+
− t
auatu
2
2
1tatuS
→→→+= = jt
at
uit
aut
−+
− 22
22
1
222
1
2
3 = Sx i + Sy j
The magnitude of the displacement = 22ySSx +
METHOD - II
This can be solved by vector addition method also. It may be noted here that →u t
will be along the direction of →u , ta
→ and
2
2
1ta
→ will be along the direction of
→a
tauV→→→
+=
Since the angle between →u and ta
→ is 1650, the magnitude of the velocity is
( ) ( ) 022 1652 cosatuatu ++
2
2
1tatuS
→→→+=
Since the angle between →u t and
2
2
1ta
→ is 1650, the magnitude of the
displacement is ( ) ( ) 022
22 1652
12
2
1cosatutatut
+
+
KINEMATICAL EQUATIONS IN RELATIVE FORM ( CONSTANT ACCELERATION ) :
030
045a
x
y
u
030
→
u
→V
ta→
045
2
2
1ta
→
030
tu→
→
S
045
11 Kinematics_______________________________________________________________________________
When two particles A and B move simultaneously with initial velocities →u A and Bu
→, at any time 't'
→→
= AAB VV - →
BV ; →→→
−= BAAB SSS ; →→→
−= BAAB aaa
→
ABU = BA UU→→
−
→
ABV = tau ABAB
→→
+
2
2
1tatuS ABABAB
→→→
+=
where →
ABX means parameter X of A with respect to B.
Similarly if →r is the position coordinate at time 't' and
→
0r is the initial position coordinate at time 't' = 0,
2
10 ++=
→→→turr AAA 2taA
→
2
10 ++=
→→→turr BBB 2taB
→
→→= ABAB rr 0 + 2
2
1tatu ABAB
→→+ gives the position coordinate of A with respect to B at any time.
→
ABr gives the distance between A and B at any time 't'.
Illustration – 9 :
A loose bolt falls from the roof of a lift of height 'h' moving vertically upward with acceleration 'a'. Find the time
taken by the bolt to reach the floor of the lift and the velocity of impact.
Solution :
jhS b −=→
l as the bolt travels a distance 'h' down wards before hitting the floor
ll
→→→−= aaa bb = (-g j ) - (a j ) = - (g + a) j
ll
→→→−= uuu bb = ju - ju = 0 as they have the same initial velocity upwards
2
2
1tatuS bbb lll
→→→+=
12 Kinematics_______________________________________________________________________________
- h j = 0 - 2
1 (a + g) t2 j ⇒ t =
ga
h
+
2
Velocity of impact is nothing but the relative velocity of the bolt with respect to the lift
Vimpact = tauV bbb lll
→→→+=
= - (a+g) j ga
h
+
2 = ( )gah +− 2 j
Illustration – 10 :
Two particles A and B move on a horizontal surface with
constant velocities as shown in the figure. If the initial distance
of separation between them is 10 m at t=0, find the distance
between them at t = 2 sec
Solution :
Distance between them = ABr→
Taking the origin at the initial position of A
2
02
1taturr ABABABAB
→→→→++=
ir AB 100 −=→
( ) ( )jcosicosjcosicosUUu BAAB0000 301060104521045210 +−−=−=
→→→
= 5 i - 18.7 j
0=−=→→→
BAAB aaa
( ) ( )j.iir AB 718510 −+−=→
t
At t = 2 sec , j.r AB 437−=→
and ABr→
= 37.4 units
DISPLACEMENT - TIME GRAPHS :
m10060
s/muB 10=
045
s/muA 210=
A B
13 Kinematics_______________________________________________________________________________
The displacement is plotted along 'y' axis and the time along 'x' axis. The slope of the curve dt
dS gives the
instantaneous velocity at that point. The average slope between two points t
S
∆
∆ gives the average velocity
between these points.
Rate of change of slope gives the acceleration. If the slope is positive and decreases with time, the particle
is under retardation. If the slope is positive and increases with time, the particle is under acceleration, constant
slope implies zero acceleration.
Illustration – 11 :
The displacement - time graph of a particle moving along a straight line is
given below. Find
a) the time at which the velocity is zero
b) the velocity at time t = 1 sec
c) the average velocity between t = 2 sec and t = 4 sec
Solution :
a) Velocity is zero when the slopoe is zero which happens at t = 2 sec
b) Since any point (x,t) lies on the circle of radius 2 m and centre (2,0),
(x-0)2 + (t - 2)2 = 22 ⇒ x = ( )± −− 224 t
velocity is given by the slope dt
dx = V
V = dt
d ( )
( )( )( )
−−
−−±=
−− 22242
124
2
2 tt
t = +
3
1
Since the slope is +ve between t = 0 and t = 2, v = 3
1 m/s
c) Average velocity = time
ntDisplaceme =
24
2
−
−O = -1 m/s
VELOCITY - TIME GRAPH :
If velocity is plotted on 'y' axis and time is plotted on x axis, the slope of the curve at any point dt
dv given
instantaneous acceleration. The average slope between two points t
v
∆
∆
0
m2x
2 4 t
circleSemi
14 Kinematics_______________________________________________________________________________
gives average acceleration. The total area between the curve and the time axis gives distance where as algebraic
sum of the areas gives displacement.
Distance = A1 + A2 + A3
Displacement = A1 - A2 + A3
The nature of acceleration can be found from the rate of change of slope
Illustration – 12 :
The velocity time graph of a particle moving along a straight line has the form of a parabola
v = (t2 - 6t + 8) m/s . Find
a) the distance travelled between t = 0 second t = 3 sec
b) the velocity of the particle when the acceleration is zero
c) the acceleration of the particle when the velocity is zero
d) the velocity of the particle when the acceleration is zero
Solution :
a) Distance = area OAB + area BCF which can be obtained by the
method of integration.
Since at the points B and D, velocity becomes zero t2 – 6t + 8 =
0
⇒ t = 2 sec and 4sec
Since F is in between B and D, the time corresponding to F is
2
42 + = 3 sec. Similarly A corresponds to t = 0 and E corresponds to t = 6 sec
Area OAB = A1 = ( )∫∫
+−=+−=
2
0
2
0
232
2
0
82
6
386 t
ttdtttdtV =
3
20m
Area BCF = A2 = - mttt
dtV3
28
2
6
3
233
2
=
+−−=∫
Distance = m3
22
3
2
3
20=+
V1A
2A
3At
V
EA
1AO
B F D 4A
3A2A t
C
15 Kinematics_______________________________________________________________________________
b) displacement between t = 3 sec and t = 6 sec = A4 - A3 = A1 - A2 = 3
2
3
20− = 6m
c) a = dt
dv= 2t - 6
When V = 0 ; t = 2sec and 4 sec
⇒ a = 2(2) - 6 and 2(4) - 6
= - 2 m/sec2 and 2 m/sec2
d) When a = 0, 2t - 6 = 0 and t = 3 sec
V = 32 - 6(3) + 8 = -1 m/sec
PROJECTILE MOTION :
At the top most point Vy = 0 and
Vx = u cosθ
From Vy = uy + ay t, 0 = usinθ - gt
⇒ t = g
sinu θ
Time of flight = 2t = g
sinu θ2
From 22yy uV − = 2ay Sy, 0 - (u sinθ)2 = 2 (-g) H and H =
g
sinu
2
22 θ
Range = (Time of flight) (horizontal velocity) = ( )g
sinucosu
g
sinu θ=θ
θ 22 2
Range is maximum when θ = 450 and Rmax = g
u2
H
R =
θ
θθ
g
sinu
g
cossinu
2
2
22
2
= 4 cot θ
The velocity of the particle at any time 't' is given by jViVV yx +=→
→V = (ucosθ) i + (usinθ - gt) j
The magnitude of the velocity = ( ) ( )22 gtsinucosu −θ+θ
R
H
T
u
θ
16 Kinematics_______________________________________________________________________________
If α is the angle made by the velocity at any time 't' with the horizontal,
Tan α = θ
−θ
cosu
gtsinu
Taking the origin at the point of projection, the 'x' and 'y' coordinates at any time 't' are given by
x = u cosθ t and y = usinθt 2
1− gt2
Eliminating 't' from x and y
y = u sinθ
θcosu
x
2
1− g
2
θcosu
x
= x tanθ - θ22
2
2 cosu
gx which is the equation of a parabola.
It may be noted here that the velocity of the projectile will be always tangential to its path. The equations
of projectile motion derived above are valid only for constant acceleration due to gravity 'g'.
Illustration – 13 :
A particle is projected from the horizontal at an inclination of 600 with an initial velocity 20 m/sec.
Assuming g = 10 m/sec2 find a) the time at which the energy becomes three fourths kinetic and one fourth
potential b) the angle made by the velocity at that time with the horizontal c) the x and y coordinates of the
particle taking the origin at the point of projection.
Solution :
a) Let V be the velocity when the given condition is fulfilled 2
1 mV2 =
4
3 (
2
1 mu2)
V = 2
3u = 10 3 m/sec
( ) jgtsinuicosuV −θ+θ=→
= 20 cos 600 i + ( )tsin 106020 0 − j = 10 i + ( ) jt10310 −
→V = 10 3 ⇒ 102 + ( )210310 t− = ( )2310
Solving t = ( )23 ± sec
t = 23 − while rising up and t = 23 + while coming down
17 Kinematics_______________________________________________________________________________
b) Tanα = θ
−θ
cosu
gtsinu
= ( )
10
2310310 ±− = + 2
c) x = ucos θ t
= 10 ( )23 ± = 10 ( ) ( )morm 231023 +−
and y = u sinθ t - 2
1 gt2
= 10 3 ( )23 ± - 5 ( ) m5232
=±
PROJECTILE MOTION ON AN INCLINED PLANE :
Let α be the inclination of the plane and the particle is
projected at an angle θ with the inclined plane. It is convenient
to take the reference frame with x' along the plane and y'
perpendicular to the plane. gcosα will be the component of
the acceleration along the downward perpendicular to the plane
and g sin α will be the component of the acceleration along the downward direction of the inclined plane.
Along the plane, the kinematical equations take the form
tauV 'x'x'x +=
⇒ tsingcosuV 'x α−θ=
2
2
1tat'uS 'xx'x +=
⇒ 'xS = ucosθt - 2
1 gsinα t2
2
'xV - 'x'x'x Sau 22
= ⇒ 2
'xV - (ucosθ)2 = 2 (-g sinα) 'xS
Similarly perpendicular to the plane, the kinematical equations take the form
tauV 'y'y'y += ⇒ tcosgsinuV 'y α−θ=
2
2
1tatuS 'y'y'y += ⇒ 2
2
1tcosgtsinuS 'y α−θ=
'y'y'y'y SauV 222 =− ⇒ ( ) ( ) 'y'y ScosgsinuV α−=θ− 222
Here it may be noted that,
αθ
'y y
u 'x
αcosgαsing
x
18 Kinematics_______________________________________________________________________________
When the particle strikes the inclined plane 'yS = 0
When the particle strikes the inclined plane perpendicular to it, 0='yS and 0='xV
When particle strikes the inclined plane horizontally 0='yS and yV = 0
Illustration – 14 :
From the foot of an inclined plane of inclination α, a projectile
is shot at an angle β with the inclined plane. Find the relation
between α and β if the projectile strikes the inclined plane
a) perpendicular to the plane
b) horizontally
Solution :
a) Since the particle strkes the plane perpendicularly 0='yS and 0='xV
u sinβ t - 2
1 g cosα t2 = 0 and u cosβ - g sinα t = 0
⇒ t = α
β
cosg
sinu2 and t =
α
β
sing
cosu
⇒ α
β=
α
β
sing
cosu
cosg
sinu2 ⇒ 2 Tan β = cot α
b) Since the particle strikes the plane horizontally 'yS = 0 and 0=yV
u sinβ t - 2
1 g cos α t2 = 0 and u sin (α + β) - gt = 0
t = α
β
cosg
sinu2 =
( )g
sinu β+α ⇒
( )g
sinu
cosg
sinu β+α=
α
β2 ⇒
α
β
cos
sin2 = sin (α + β)
CIRCULAR MOTION :
When a particle moves in a circle of radius R with constant speed V, its called uniform circular motion.
αβ
'y
u 'x
Vθ
V
θV
V
V∆
19 Kinematics_______________________________________________________________________________
When the particle covers θ, the direction of velocity also changes by θ without change in magnitude.
Change in velocity ∆V will be towards the centre of curvature of the circular path which causes centripetal
acceleration. θ is called the angular position (or) angular displacement.
Centripetal acceleration, ra = t
V
∆
∆
The rate of change of angular position is known as angular velocity (ω)
Time period of circular motion T = V
Rπ2
In the same time the particle covers an angle 2 π from which angular velocity can be found as
ω = T
π2 =
R
V
R
V=
π
π
2
2
⇒ ∆t = ω
θ =
V
Rθ and ∆ V = θ−+ cosVVV 222 2 = 2V sin
2
θ
When θ is small sinθ ~ θ ⇒ ∆ V = Vθ
centripetal acceleration = t
V
∆
∆ =
( )
θ
θ
V
R
V =
R
V2
When speed of the particle continuously changes with time, the tangential acceleration is given by
ta = dt
dV
The rate of change of angular velocity is called the angular acceleration (α)
since ra and ta are perpendicular to each other, the resultant acceleration is given by a =
22tr aa +
Angle made by the resultant with radius vector Tanθ = r
t
a
a
20 Kinematics_______________________________________________________________________________
Illustration – 15 :
The speed of a particle in circular motion of radius R is given by V = Rt2. Find the time at which the radial and the
tangential accelerations are equal and the distance traveled by the particle upto that moment.
Solution :
tr aa =
dt
dV
R
V=
2
= 2Rt ⇒ t = 3
1
2
Distance travelled = ∫3
1
2
0
dtV =
3
1
23
3
Rt =
3
2R
RADIUS OF CURVATURE
When a particle is moving in a plane R
Va r
2
= where V is the instantaneous velocity and R is the radius
of curvature at that point.
Radius of curvature = ra
V2
If the path of the particle is given by y = f(x), radius of curvature can also be found from the formula R =
2
2
2
32
1
dx
yd
dx
dy
+
Illustration – 16 :
A particle is projected with initial velocity 'u' at angle θ with the horizontal. Find the radius of curvature at
a) point of projection b) the top most point
Solution :
a) at the point of projection P, V = u and ra = g cosθ
θ
u
p
T
21 Kinematics_______________________________________________________________________________
⇒ R = ra
V2
= θcosg
u 2
b) at the topmost point T, V = ucosθ and ga r =
⇒ R = ra
V2
= g
cosu θ22
SHORTEST DISTANCE OF APPROACH :
When two particles A and B are moving simultaneously, their position coordinates at any time 't' are given
by (when the accelerations are uniform) 20
2
1taturr AAAA
→→→→
++=
and 20
2
1taturr BBBB
→→→→
++=
The distance between them at any time 't', S = ABr→
Where 20
2
1taturr ABABABAB
→→→→
++=
The distance between them becomes minimum when dt
dS=0 from which the time at which it becomes
minimum can be found. Substituting the value of time so obtained in ABr→
, S min can be found.
Illustration – 17 :
Two ships A and B move with constant velocities as
shown in the figure. Find the closest distance of
approach between them
Solution :
jr A 100 =→
ir B 200 =→
jcosicosV A 30206020 0 −=→
jcosicosVB 4521045210 +=→
ji 31010 −= = 10 i + 10 j
A
km10
o30
North
kmphVA 20=
kmphVB 210=
East
045
O km20 B
22 Kinematics_______________________________________________________________________________
0=→
Aa 0=→
Ba
tVrr AAA
→→→
+= 0 tVrr BBB
→→→
+= 0
= 10 t i + ( ) jt31010 − = (20 + 10 t) i + 10 t j
BAAB rrr→→→
−= = - 20 i + ( ) jtt 1031010 −−
S = ABr→
= ( ) ( )22 103101020 tt −−+−
When the distance between A and B is minimum dt
dS = 0
( ) ( )
−−+−22
1031010202
1
tt
( ) ( )( )1031010310102 −−−− tt = 0
10 - 10 3 t - 10 t = 0 t =
+ 31
1 hr
Substituting this value of time in the expression for S,
S min = 20 km
CYCLIC MOVEMENT OF PARTICLES :
When three or more particles located at the vertices of a polygon of side l move with constant speed V such
that particle 1 moves always towards particle 2 and particle 2 moves always towards 3 particle etc., they meet at the
centre of the polygon following identical curved paths.
Time of meeting = approachVelocityof
seperationInitial
Velocity of approach is the component of the relative velocity along the line joining the particles.
23 Kinematics_______________________________________________________________________________
Illustration – 18 :
Six particles located at the six vertices of a hexagon of side l move with constant speeds V such that each particle
always targets the particle in front if it. Find the time of meeting and the distance travelled by each particle before
they meet
Solution :
t = approachofVelocity
seperationInitial
= 060cosVV −
l =
V
l2
Since they move with constant speed V, the distance travelled by each particle
in time t = V
l2 is d = Vt = V
V
l2 = 2 l
RIVER PROBLEMS :
If rV→
is the velocity of the river and bV→
is the velocity
of the boat with respect to still water, the resultant
velocity of the boat RV→
= rb VV→→
+
Only the perpendicular component of the resultant
velocity helps in crossing the river. Time of
crossing, t =θcosV
w
b where 'w' is the width of the river.
The boat crosses the river in the least time when θ = 0
The parallel component of the resultant velocity determines the drift.
Drift is the displacement of the boat parallel to the river by the time the boat crosses the river
Drift , x = ( )θ− sinVV br
θcosV
w
b
Zero drift is possible only when Vr = Vb sinθ. When Vr > Vb zero drift is not possible.
Illustration – 19 :
A river of width 100 m is flowing towards East with a velocity of 5 m/s. A boat which can move with a speed of 20
m/s with respect to still water starts from a point on the South bank to reach a directly opposite point on the North
bank. If a wind is blowing towards North East with a velocity of 5 2 m/s, find the time of crossing and the angle
at which the boat must be rowed.
V
V
V
V
V
060
B
θ
rV
A
C•
bV
24 Kinematics_______________________________________________________________________________
Solution:
jcosisinV b θ+θ−=→
2020
rV→
= 5 i
→
V w = 5 2 cos 45 i + 5 2 cos 45 j = 5 i + 5 j
RV→
= Resultant velocity of the boat = bV→
+ rV→
+ wV→
= ( - 20 sin θ + 5 + 5) i + (20 cos θ + 5) j
For reaching directly opposite point, the component of the resultant velocity parallel to the river must be
zero
- 20 sinθ + 10 = 0 ⇒ sinθ = 2
1 and θ = 300
Since time of crossing depends only on the perpendicular component of the resultant velocity.
t = 520 +θcos
w =
53020
1000 +cos
= 4.48 sec
WORKED OUT OBJECTIVE PROBLEMS
EXAMPLE : 01
A point moves along 'x' axis. Its position at time 't' is given by x2 = t 2 + 1. Its acceleration at time 't' is
A) 3
1
x B)
2
11
xx− C)
2x
t− D)
3
2
x
t−
Solution :
x = 12 +t ; dt
dx =
12
12 +t
(2t) = 12 +t
t
a = 2
2
xt
xd =
22
2
2
1
212
1
+
+−
+
t
)t(t
tt
=
( )32 1
1
+t
= 3
1
x
rVwV
45θ
bV
25 Kinematics_______________________________________________________________________________
EXAMPLE : 02
A body thrown vertically up from the ground passes the height 10.2m twice at an interval of 10 sec. Its
initial velocity was (g = 10 m/s2)
A) 52 m/s B) 26 m/s C) 35 m/s D) 60 m/s
Solution :
Displacement is same in both cases
s = ut + 1/2 at2
10.2 = ut - 2
1 (10) t2 ⇒ t =
10
2042 −± uu
t1 = 10
2042 −− uu and t2 =
10
2042 −+ uu ⇒ ∆t = t2 - t1 = 10 sec
2042 −u = 50 ⇒ u2 = 2500 + 204 ⇒ u = 52 m/s
EXAMPLE : 03
A car starts from rest moving along a line, first with acceleration a= 2 m/s2, then uniformly and finally
decelerating at the same rate and comes to rest. The total time of motion is 10 sec. The average speed
during this time is 3.2 m/s. How long does the car move uniformly
A) 4 sec B) 6 sec C) 5 sec D) 3 sec
Solution :
Let the car accelerate for time 't' and move uniformly with v = at for time t1 Since the magnitudes of
acceleration and deceleration are same, the time of deceleration is also 't'.
t + t 1 + t = 10 sec
Average speed = time
Distance =
( )
10
2
1
2
1 21
2
++
attatat
= 3.2
2t2 + 2tt1 = 32 ⇒ 2 322
102
2
101
12
1 =
−+
−t
tt
Solving t1 = 6 sec This problem can be solved using velocity time graph also.
EXAMPLE : 04
A particle has an initial velocity of ( )ji 43 + m/s and a constant acceleration ( )ji 34 − m/s2. Its speed after
1 sec will be equal to
26 Kinematics_______________________________________________________________________________
A) zero B) 10 m/s C) 5 2 m/s D) 25 m/s
Solution :
tauV→→→
+= = ( ) ( )jiji 3443 −++ (1) = 7 i - j
Speed = magnitude of →
V = 22 17 + = 5 2 m/s
EXAMPLE : 05
An aeroplane flies along a straight line from A to B with air speed V and back again with the same air speed.
If the distance between A and B is l and a steady wind blows perpendicular to AB with speed u, the total time
taken for the round trip is
A) V
l2 B)
22
2
uV +
l C)
22
2
uV
Vl D)
22
2
uV −
l
Solution :
The resultant velocity of the plane must be along AB during forward journey.
t1 = 22 uVVR −
=ll
During return journey, the resultant velocity of the plane must be along BA
t2 = 22 uVVR −
=ll
Total time t = t1 + t2 = 22
2
uV −
l
EXAMPLE : 06
A particle is thrown with a speed 'u' at an angle θ with the horizontal. When the particle makes an angle φ
with the horizontal its speed changes to V. Then
A) V = u cosθ B) V = ucosθ cosφ C) V = u cosθ secφ D) V = usecθ cosφ
Solution :
Since the horizontal component of the velocity of a projectile always remains constant
u cosθ=V cosφ ⇒ V=ucosθ secφ
A
RV
B
V
u
RV
V
A
Bu
27 Kinematics_______________________________________________________________________________
EXAMPLE : 07
Two shells are fired from a cannon with same speed at angle α and β respectively with the horizontal. The
time interval between the shots is T. They collide in mid air after time 't' from the first shot. Which of the
following conditions must be satisfied.
A) α > β B) t cosα = (t -T) cos β
C) (t-T) cosα=cosβ D) (usinα)t - 2
1 gt2 =(usinβ) (t-T)-
2
1g(t-T)2
Solution :
When they collide, their 'x' and 'y' components must be same
ucosα t = u cosβ (t-T) ⇒ cosα t = cos β(t-T)
(usinα) t - 2
1 gt2 = (usinβ) (t-T) -
2
1 g (t-T)2
Since cosα = cos β
−
t
T1 and T < t
cos α < cosβ and α > β
EXAMPLE : 08
A particle is projected from a point 'p' with velocity 5 2 m/s perpendicular to the surface hollow right
angle cone whose axis is vertical. It collides at point Q normally on the inner surface. The time of flight of
the particle is
A) 1 sec B) 2 sec C) 2 2 sec D) 2 sec
Solution :
It can be seen from the diagram that →
V becomes perpendicular to →
u .
→
u = ucos450 i + u sin45 j
→
V = →
u + →
a t = (ucos 45 i + usin45 j ) - (gt) j
When →
V becomes perpendicular to →
u , →V .
→u = 0
u2 cos2 45 + u2 sin2 45 - (usin45) gt = 0 ⇒ t = 45sing
u = 1 sec
EXAMPLE : 09
28 Kinematics_______________________________________________________________________________
A man walking Eastward at 5 m/s observes that the wind is blowing from the North. On doubling his speed
Eastward he observes that the wind is blowing from North East. The velocity of the wind is
A) (5i+5j) m/s B) (5i - 5j) m/s C) (-5i +5j) m/s D) (-5i - 5j) m/s
Solution :
let jViVV w 21 +=→
In the first case mwwm VVV→→→
−= = ( )jViV 21 + - ( )i5
Since no component along East is observed V1 - 5 = 0 ⇒ V1 = 5 m/s
In the second case
mwwm VVV→→→
−= = (V1 i + V2 j ) - (10 i ) = ( ) jViV 21 10 +−
Since the wind is observed from North East the components along North and East must be same V1 - 10
= V2 ⇒ V2 = - 5 m/s
→
V w = (5i - 5j) m/s
EXAMPLE :10
From a lift moving upward with uniform acceleration 'a', a man throws a ball vertically upwards with a
velocity V relative to the lift. The time after which it comes back to the man is
A) ag
V
−
2 B)
ag
V
+ C)
ag
V
+
2 D)
22
2
ag
Vg
−
Solution :
Since the velocity of the ball is given relative to the lift blV→
= V j
When the ball comes back to the man, its displacement relative to the lift is zero blS→
= 0
bla→
= lb aa→→
− = (-g) j - a j = - (g + a) j
Applying S = ut + 1/2 at2 in relative form
blbl VS→→
= t + 2
1 bla
→
t2
0 = ( )jVt + 2
1 ( )( )jag +− t2 ⇒ t =
ag
V
+
2
29 Kinematics_______________________________________________________________________________
ASSIGNMENT
SINGLE ANSWER TYPE QUESTIONS
LEVEL – I
1. The greatest acceleration or deceleration that a train may have is a. The minimum time in which the train
can go from one station to the next at a distance S is
A) a
S B)
a
S2 C) 2
a
s D)
a2
s
2. A car accelerates from rest at a constant rate α for sometime and attains a velocity of 20 m/s. Afterwards it
decelerates with a constant rate α/2 and comes to a halt. If the total time taken is 10s, the distance travelled
by the car is
A) 200m B) 100m C) 10m D) 20m
3. A particle starts from the position of rest under a constant acceleration. It travels a distance x in the first 10
seconds and distance y in the next 20 seconds. Then
A) y = x B) y = 2x C) y = 8x D) y = 4x
4. A body is projected vertically upwards. If t1 and t2 be the times at which it is at height h above the point of
projection while ascending and descending respectively, then h is
A) 21ttg2
1 B) g t1 t2 C) 2 g t1 t2 D) 4 g t1 t2
5. From a 20m high tower one ball is thrown upward with speed of 10m/s and another is thrown vertically
downward at the same speed simultaneously. The time difference of their reaching the ground will be (take
g = 10m/s2)
A) 12s B) 6s C) 2s D) 1s
6. A particle X moving with a constant velocity u crosses a point O. At the same instant another particle F
starts from rest from O with a constant acceleration a. The maximum separation between them before they
meet is
A) a2
u2
B) a
u2
C) a
u2 2
D) a4
u2
7. A bird flies in straight line for 4s with a velocity v = (2t-4) m/s. What is the distance covered by the bird in
returning to the place from where it started its journey ?
A) 0 B) 8m C) 4m D) 2m
8. A ball is thrown vertically upwards. Which of the following plots represents the speed-time graph of the
ball during its flight if the air resistance is not ignored
30 Kinematics_______________________________________________________________________________
A) B) C) D)
9. If a ball is thrown vertically upwards with speed 'u', the distance covered during the last t second of its
ascent is
A) (u+gt)s B) ut C) 2
1gt2 D) ut -
2
1gt2
10. A particle has an initial velocity of 9m/s due east and a constant acceleration of 2m/s2 due west. The
distance covered by the particle in the fifth second of its motion is
A) 0 B) 0.5m C) 2m D) none of these
11. Two particles are projected simultaneously in the same vertical plane from the same point, with different
speeds u1 and u2, making angles θ1 and θ2 respectively with the horizontal , such that u1 cosθ1 = u2cosθ2.
The path followed by one, as seen by the other (as long as both are in flight), is :
A) a horizontal straight line B) a vertical straight line C) a parabola
D) a straight line making an angle |θ1 - θ2| with the horizontal
12. A particle starts from the origin of coordinates at time t = 0 and moves in the xy plane with a constant
acceleration α in the y-direction. Its equation of motion is y =βx2. Its velocity component in the x-direction
is
A) variable B) β
α2 C)
β
α
2 D)
β
α
2
13. A train starts from station A with uniform acceleration α for some distance and then goes with uniform
retardation β for some more distance to come to rest at station B. The distance between station A and B is 4
km and the train takes 4 minute to complete this journey. If α and β are in km (min)-2 then
A) 211
=β
+α
B) 411
=β
+α
C) 2
111=
β+
α D)
4
111=
β+
α
14. The driver of a train moving with a speed v1 sights another train at a distance d, ahead of him moving in the
same direction with a slower speed v2. He applies the breaks and gives a constant de-acceleration 'a' to his
train. For no collision, d is
A) = ( )
a2
vv 221 −
B) > ( )
a2
vv 221 −
C) < ( )
a2
vv 221 −
D) < a2
vv 21 −
15. In the case of a moving body, pick the correct statement
A) if speed changes with change in direction, velocity does not change
31 Kinematics_______________________________________________________________________________
B) if velocity changes, speed may or may not change but acceleration does change
C) if velocity changes, speed also changes with same acceleration
D) if speed changes without change in direction, the velocity may remain constant.
16. Particle 1 is in one dimensional motion with uniform velocity whereas particle 2 is accelerating in a straight
line. The graph representing path of 2 with respect to 1 is
A) B) C) D)
17. Person A walking along a road at 3ms-1 sees another person B walking on another road at right angles to his
road. Velocity of B is 4ms-1when he is 10m off. They are nearest to each other when person A has covered
a distance of
A) 3.6m B) 8m C) 6.3m D) 0.8m
18. The given graph shows the variation of velocity with displacement. Which one of the
graphs given below correctly represents the variation of acceleration with displacement.
A) B) C) D)
19. v2 versus s-graph of a particle moving in a straight line is as shown in figure.
From the graph some conclusions are drawn. State which statement is wrong :
A) the given graph shows a uniformly accelerated motion.
B) initial velocity of particle is zero
C) corresponding s-t graph will be a parabola
D) none of the above
20. A body dropped from the top of the tower covers a distance 7x in the last second of its journey, where x is
the distance covered in first second. How much time does it take to reach the ground ?
A) 3s B) 4s C) 5s D) 6s
21. A body is projected with a velocity u. It passes through a certain point above the ground after t1 sec. The
time interval after which the body passes through the same point during the return journey is
A)
− 2
1tg
u B) 2
− 1tg
u C) 3
− 1
2
tg
u D) 3
− 12
2
tg
u
22. The area of the acceleration-displacement curve of a body gives :
A) impulse B) change in momentum per unit mass
32 Kinematics_______________________________________________________________________________
C) change in KE per unit mass D) total change in energy
23. A body thrown vertically up from the ground passes the height 10.2m twice at an interval of 10s. What was
its initial velocity (g = 10m/s2)
A) 52m/s B) 26 m/s C) 35 m/s D) 60 m/s
24. An insect crawls a distance of 4m along north in 10 seconds and then a distance of 3m along east in 5
seconds. The average velocity of the insect is :
A) 7/15 m/sec B) 1/5 m/sec C) 1/3 m/sec D) 4/5 m/sec
25. A particle returns to the starting point after 10s. If the rate of change of velocity during the motion is
constant in magnitude, then its location after 7 seconds will be same as that after :
A) 1 second B) 2 seconds C) 3 seconds 4) 4 second
26. Two particles P and Q simultaneously start moving from point A with velocities 15m/s and 20m/s
respectively. The two particles move with accelerations equal in magnitude but opposite in direction. When
P overtakes Q at B then its velocity is 30m/s. The velocity of Q at point B will be
A) 30m/s B) 5m/s C) 20m/s D) 15m/s
27. Let aandv denote the velocity and acceleration respectively of a particle in one dimensional motion:
A) the speed of the particle decreases when 0av <•
B) the speed of the particle increases when 0av >•
C) the speed of the particle increases when 0av =•
D) the speed of the particle decreases when av <
28. The displacement-time graph of a moving particle is shown in figure. The
instantaneous velocity of the particle is negative at the point :
A) C B) D
C) E D) F
29. A ball A is thrown vertically upward with speed u. At the same instant
another ball B is released from rest at height h. At time t the speed of A relative to B is :
A) u B) u-2gt C) gh2u 2 − D) u-gt
30. The velocity of a projectile at any instant is u making an angle α to the horizon. The time after which it will
be moving at right angle to this direction is :
A) g
eccosu α B)
g
sinu2 α C)
g
cosu α D)
g
tanu α
31. A ball is thrown up with a certain velocity at an angle θ to the horizontal. The kinetic energy KE of the ball
varies with horizontal displacement x as :
33 Kinematics_______________________________________________________________________________
A) B) C) D)
32. A ball of mass m is projected from the ground with an initial velocity u making an angle of θ with the
horizontal. What is the change in velocity between the point of projection and the highest point :
A) u cos2 θ downward B) u cos θ upward C) u sin2θ upward D) u sin θ downward
33. Read the following statements and state whether they are True or False.
i) If u and a both are negative, motion is only retarded.
ii) If u is negative but a is positive, then displacement of the particle can never be positive.
iii) If u is positive but a is negative particle comes to rest for a moment at some time but if u is negative and
a positive it never comes to rest.
A) T, T, T B) F, F, F C) T, F, T D) F, T, F
34. Suppose in the absence of air resistance, R = OB, H = AC, t1 = tOA and t2 = tOB. If
air resistance is taken into consideration and the corresponding values are R', H',
'1t and t2' then
A) R' < R, H' < H, '1t > t1 and '
2t > t2
B) R' < R, H' < H, '1t > t1 and '
2t < t2
C) R' < R, H' > H, '1t > t1 and '
2t < t2
D) R' < R, H' < H, '1t < t1 and '
2t > t2
35. Speed of a particle moving in a circle varies with time as, v = 2t. Then :
A) angle between velocity vector and acceleration vector is increasing with time.
B) a is constant while ar is increasing with time.
C) both A and B are correct D) both A and B are wrong.
36. Initial velocity and acceleration of two particles are as shown in fig. Assuming the shown direction as the
positive, vBA versus time graph is as :
A) B) C) D)
37. A graph is plotted between velocity (v) and displacement (s) of particle moving in a straight line. Here v is
plotted along y-axis and 's' along x-axis. Choose the correct option.
A) slope of this graph at any point always gives us the ratio of velocity and displacement at that point.7
34 Kinematics_______________________________________________________________________________
B) slope represents a/v under all the conditions. (a = acceleration)
C) both A and B are correct D) both A and B are wrong.
38. In a projectile motion if a person wants to increase the maximum height to 2 times but simultaneously want
to decrease the range same number of time. He can achieve it by increasing tan
of angle of projection by ....... times.
A) 2 B) 4 C) 3 D) 2
39. The velocity of a particle moving in a straight line varies with time in such a
manner that v versus t graph is represented by one half of an ellipse. The
maximum velocity is νm and total time of motion is t0
i) Average velocity of particle is π/4 νm ii) Such motion can not be realized in practical terms
A) Only (i) is correct B) Only (ii) is correct
C) Both (i) and (ii) are correct D) Both (i) and (ii) are wrong
40. Starting from rest, a particle rotates in a circle of radius R = 2 m with an angular acceleration α = π/4 rad/ s2.
The magnitude of average velocity of the particle over the time it rotates quarter circle is
A) 1.5 m/s B) 2 m/s C) 1 m/s D) 1.25m/s
41. In a car race car A takes t0 time less to finish than car B and passes the finishing point with a velocity v0
more than car B. The cars start from rest and travel with constant accelerations a1 and a2. Then the ratio ν0/t0
is equal to
A) 2
21
a
a B)
2
aa 21 + C) 21 aa D)
1
22
a
a
42. A rod of length 1 leans by its upper end against a smooth vertical wall, while its
other end leans against the floor. The end that leans against the wall moves
uniformly downward. Then
A) The other end also moves uniformly
B) The speed of other end goes on decreasing
C) The speed of other end goes on increasing
D) The speed of other end first decreases and then increases
43. A particle is moving along a circular path of radius 5 m and with uniform speed 5 m/s. What will be the
average acceleration when the particle completes half revolution?
A) zero B) π m/s2 C) 10 πm/s2 D) 10/πm/s2
44. The velocity displacement graph of a particle moving along a straight line is shown
The most suitable acceleration-displacement graph will be
35 Kinematics_______________________________________________________________________________
A) B)
C) D)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
C B C A C A B C C B B D A B B A A A B B
21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
B C A C C B A C A A C D B D C A B B C C
41 42 43 44
C A
LEVEL – II
1. A particle starts from rest at time t = 0 and moves on a straight line
with an acceleration which varies with time as shown in fig. The
speed of the particle will be maximum after how many seconds
A) 4s B) 6s
C) 8s D) 10s
2. Due to air a falling body faces a resistive force proportional to square of velocity v, consequently its
effective downward acceleration is reduced and is given by a = g - kv2 where k = 0.002m-1. The terminal
velocity of the falling body is
A) 49m/s B) 70m/s C) 9.8m/s D) 98m/s
3. A balloon is rising with a constant acceleration of 2m/s2. At a certain instant when the balloon was moving
with a velocity of 4m/s, a stone was dropped from it in a region where g = 10m/s2. The velocity and
acceleration of stone as it comes out from the balloon are respectively.
A) 0, 10m/s2 B) 4m/s, 8m/s2 C) 4m/s, 12m/s2 D) 4m/s, 10m/s2
4. A stone is thrown vertically up from the top of a tower with some initial velocity and it arrives on the
ground after t1 seconds. Now if the same stone is thrown vertically down from the top of the same tower
36 Kinematics_______________________________________________________________________________
with the same initial velocity, it arrives on ground after t2 seconds. How much time will the stone take to
reach the ground if it is dropped from the same tower ?
A) 2
tt 21 + B)
2
tt 21 − C) 21 tt + D) 21tt
5. Two particles A and B start from the same point and slide down through straight smooth planes inclined at
300 and 600 to the vertical and in the same vertical plane and on the same side of vertical drawn from the
starting point. The acceleration of B with respect to A is
A) g/2 in vertical direction B) g 2/3 at 450 to vertical
C) g/ 3 at 600 to vertical D) g in vertical direction
6. A particle starts from rest at the origin and moves along X-axis with acceleration a = 12-2t. The time after
which the particle arrives at the origin is
A) 6s B) 18s C) 12s D) 4s
7. Figure represents position (x) versus time (t) graph for the motion of a
particle. If b and c are both positive constants, which of the following
expressions best describes the acceleration (A) of the particle ?
A) a = +b
B) a = -c
C) a = b + ct
D) a = b-ct
8. Two particles instantaneously at A and B are 5m, apart and they are moving with
uniform velocities, the former towards B at 4m/s and the latter perpendicular to AB at
3m/s. They are nearest at the instant
A) 2/5s
B) 3/5s
C) 1s
D) 4/5s
9. Three particles start from the origin at the same time, one with a velocity 'u1', alone X-axis, the second
along the Y-axis with a velocity u2 and the third alogn x = y line. The velocity of third particle so that the
three may always lie on the same plane is
A) 2
uu 21 + B) 21uu C)
21
21
uu
uu
+ D)
21
21
uu
uu2
+
10. A ball is shot vertically upwards from the surface of a planet in a
distant solar system. A plot of the y versus t for the ball is shown in
fig. The magnitude of the free fall in m/s2 on the planet is
A) 4
37 Kinematics_______________________________________________________________________________
B) 8
C) 12
D) 16
11. The horizontal range of a projectile is R and the maximum height attained by it is H. A strong wind now
begins to blow in the direction of the motion of the projectile, giving it a constant horizontal acceleration =
g/2. Under the same conditions of projection, the horizontal range of the projectile will now be
A) R + 2
H B) R + H C) R +
2
H3 D) R + 2H
12. A particle moves int he xy plane with a constant acceleration g in the negative y-direction. Its equation of
motion is y = ax - bx2, where a and b are constants. Which of the following are correct ?
A) The x-component of its velocity is constant
B) at the origin, the y-component of its velocity is a b2
g
C) At the origin, its velocity makes an angle tan-1(a) with the x-axis.
D) the particle moves exactly like a projectile.
13. Two bodies are projected simultaneously from the same point, in the same vertical plane, one towards east
and other towards west with velocities 8 ms-1 and 2 ms-1 respectively. The time at which their velocities are
perpendicular to each other is
A) 2/5 s B) 5/2s C) 1/5 s D) 5 s
14. Two stones are projected so as to reach the same distance from the point of projection on a horizontal
surface. The maximum height reached by one exceeds the other by an amount equal to half the sum of the
heights attained by them. Then the angles of projection for the stones are
A) 450, 1350 B) 00, 900 C) 300, 600 D) 200, 700
15. Velocity and acceleration of a particle at some instant of time are ( )j4i3v += m/s and ( )j8i6a +−= m/s2
respectively. At the same instant particle is at origin. Maximum x-co-ordinate of particle will be
A) 1.5m B) 0.75m C) 2.25m D) 4.0m
16. a-t graph for a particle moving in a straight line is as shown in figure.
Change in velocity of the particle from t=0 to t=6s is :
A) 10m/s B) 4m/s
C) 12m/s D) 8m/s
17. Speed time graph of two cars A and B approaching towards each other is
shown in figure. Initial distance between them is 60m. The two cars will cross
each other after time.
38 Kinematics_______________________________________________________________________________
A) 2sec B) 3sec
C) 1.5sec D) 2 sec
18. The position of a particle along x-axis at time t is given by x = 2 + t - 3t2. The displacement and the
distance travelled in the interval t = 0 to t = 1 are respectively
A) 2, 2 B) -2, 2.5 C) 0, 2 D) -2, 2.16
19. The acceleration time graph of a particle moving along a straight line is as
shown in figure. At what time the particle acquires its initial velocity ?
A) 12s
B) 5s
C) 8s
D) 16s
20. A graph between the square of the velocity of a particle and the
distance s moved by the particle is shown in the figure. The
acceleration of the particle in kilometre per hour square is :
A) 2250
B) 225
C) -2250
D) -225
21. A particle starts from rest and traverses a distance l with uniform acceleration, then moves uniformly over a
further distance 2l and finally comes to rest after moving a further distance 3l under uniform retardation.
Assuming entire motion to be rectilinear motion the ratio of average speed over the journey to the
maximum speed on its way is :
A) 1/5 B) 2/5 C) 3/5 D) 4/5
22. Two stones are thrown up simultaneously with initial speeds of u1 andu2 (u2 > u1). They hit the ground after
6s and 10s respectively. Which graph in figure correctly represents the time variation of ∆x = (x2 - x1), the
relative position of the second stone with respect to the first upto t = 10s ? Assume that the stones do not
rebound after hitting the ground.
A) B) C) D)
39 Kinematics_______________________________________________________________________________
23. Figure shows the position-time (x-t) graph of the motion of two boys A and B
returning from their school O to their homes P and Q respectively. Which of the
following statements is true ?
A) A walks faster than B
B) Both A and B reach home at the same time
C) B starts for home earlier than A
D) A overtakes B on his way to home
24. A ball is projected with a velocity 20 3 m/s at angle 600 to the horizontal. The time interval after which the
velocity vector will make an angle 300 to the horizontal is (take g = 10m/s2)
A) 4sec B) 2 sec C) 1 sec D) 3 sec
25. The equation of motion of a projectile is : y = 12x - 4
3x2
Given that g = 10ms-2, what is the range of the projectile ?
A) 12m B) 16m C) 20m D) 24m
26. A projectile is thrown with an initial velocity of ( ) 1msjbia −+ . If the range of the projectile is twice the
maximum height reached by it, then :
A) a = 2b B) b=a C) b = 2a D) b = 4a
27. A particle moves along a parabolic path y = 9x2 in such a way that the x component of velocity remains
constant and has a value 1ms3
1 − . The acceleration of the particle is
A) 2msj3
1 − B) 2msj3 − C) 2msj3
2 − D) 2msj2 −
28. Two projectiles are projected with the same velocity. If one is projected at an angle of 300 and the other at
600 to the horizontal. The ratio of maximum heights reached, is :
A) 1 : 3 B) 2 : 1 C) 3 : 1 D) 1 : 4
29. A particle is projected from the ground with velocity u at angle θ with horizontal. The horizontal range,
maximum height and time of flight are R, H and T respectively. They are given by,
R = g2
sinuH,
g
2sinu 222 θ=
θ and T =
g
sinu2 θ
Now keeping u as fixed, θ is varied from 300 to 600. Then,
A) R will first increase then decrease, H will increase and T will decrease
40 Kinematics_______________________________________________________________________________
B) R will first increase then decrease, while H and T both will increase
C) R will decrease while H and T will increase
D) R will increase while H and T will decrease
30. Velocity and acceleration of a particle at some instant of time are ( )k2ji2v +−= m/s and
( )kj6ia −+= m/s2 . Then, the speed of the particle is ....... at a rate of ....... m/s2
A) increasing, 2 B) decreasing ,2 C) increasing, 4 D) decreasing, 4
31. x and y coordinates of a particle moving in xy plane at some instant are :
x = 2t2 and y = 2t2
3
The average velocity of particle in a time interval from t = 1 second to t = 2 second is :
A) ( ) s/mj5i8 + B) ( ) s/mj9i12 + C) ( ) s/mj5.4i6 + D) ( ) s/mj6i10 +
32. A particle is projected upwards with some velocity. At what height from ground should another particle be
just dropped at the same time so that both reach the ground simultaneously. Assume that first particle
reaches to a maximum height H.
A) 6H B) 8H C) 4H D) 10H
33. Particle A moves with 4m/s along positive y-axis and particle B in a circle x2 + y2 = 4(anticlockwise) with
constant angular velocity ω = 2 rad/s. At time t = 0 particle is at (2m, 0). Then :
A) magnitude of relative velocity between them at time t is 8 sin 2t
B) magnitude of relative velocity between them is maximum at t = π/4 second.
C) both A and B are correct
D) both A and B are wrong
34. An armored car 2 m long and 3 m wide is moving at 10 ms-1 when a bullet hits it in a direction making an
angle tan-1(3/4) which the length of the car as seen by a stationary observer. The bullet enters one edge of
the car at the corner and passes out at the diagonally opposite corner. Neglecting any interaction between
the car and the bullet, the time for the bullet to cross the car is
A) 0.20 s B) 0.15 s C) 0.10 s D) 0.50 s
35. The V - t graph for the rectilinear motion of a particle is represented by a parabola as shown in fig. Find the
distance traveled by the particle in time T/2.
A) 3
TV2max
B) 2
TV2max
41 Kinematics_______________________________________________________________________________
C) 3
TVmax
D) 2
TV 2max
36. A glass wind screen whose inclination with the vertical can be changed is mounted on a car. The car moves
horizontally with a speed of 2 m/s. At what angle α with the vertical should the wind screen be placed so
that the rain drops falling vertically downwards with velocity 6 m/s strike the wind screen perpendicularly.
A) tan-1 (1/3) B) tan-1 (3) C) cos-1(3) D) sin-1(1/3)
37. Two stones are thrown up simultaneously from the edge of a cliff with initial speeds v and 2 v. The relative
position of the second stone with respect to first varies with time till both the stones strike the ground as
A) Linearly B) First linearly then parabolically
C) Parabolically D) First parabolically then linearly
38. There are two values of time for which a projectile is at the same height. The sum of these two times is
equal to
A) 3T/2 B) 4T/3 C) 3T/4 D) T
39. A particle is projected from a horizontal plane with 8 2 m/s at an angle. At highest point its velocity is
found to be 8 m/s. Its range will be
A) 6.4 m B) 3.2 m C) 5 m D) 12.8 m
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
A B D D A B D D D B D A A C B B B D C C
21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39
C A B B B C D A B B C C D A C D
LEVEL - III
1. Two particles A and B are connected by a rigid rod AB. The rod slides on
perpendicular rails as shown in fig. The velocity of A to the left is 10m/s. What is
velocity of B when α = 600?
A) 10m/s
B) 5.8m/s
C) 17.3m/s D) 9.8m/s
2. In the figure , the pulley P moves to the right with a constant speed u. The
downward speed of A is vA, and the speed of B to the right is vB.
A) vB = vA
B) vB = u + vA
C) vB + u = vA
D) The two blocks have accelerations of the same magnitude
42 Kinematics_______________________________________________________________________________
3. A marble starts falling from rest on a smooth inclined plane forming an angle α with horizontal. After
covering distance 'h' the ball rebound off the plane. The distance from the impact point where the ball
rebounds for second time is
A) 8h cosα B) 8h sinα C) 2h tanα D) 4h sinα
4. From the top of a tower of height 40m, a ball is projected upwards with a speed of 20m/s at an angle of
elevation of 300. The ratio of the total time taken by the ball to hit the ground to its time of flight (time
taken to come back to the same elevation) is (take g = 10m/s2)
A) 2 : 1 B)3 : 1 C) 3 : 2 D) 1.5 : 1
5. If time taken by the projectile to reach Q is T, then PQ is equal to :
A) Tv sin θ
B) Tv cos θ
C) Tv sec θ
D) Tv tanθ
6. A particle is thrown with a speed u at an angle θ with the horizontal. When the particle makes an angle φ
with the horizontal, its speed changes to v :
A)v = u cos θ B) v = u cosθ cos φ C) v = u cosθ sec φ D) v = u sec θ cos φ
7. A stone is projected from a point on the ground so as to hit a bird on the top of a vertical pole of height h
and then attain a maximum height 2h above the ground. If at the instant of projection the bird flies away
horizontally with a uniform speed and if the stone hits the bird while descending then the ratio of the speed
of the bird to the horizontal speed of the stone is :
A) 12
2
+ B)
12
2
− C)
2
1
2
1+ D)
12
2
+
8. Shots are fired simultaneously from the top and bottom of a vertical cliff with the
elevation α = 300, β = 600 respectively and strike the object simultaneously at the same
point.
If a = 30 3 m is the horizontal distance of the object from the cliff, then the height of
the cliff is :
A) 30m B) 45m
C) 60m D) 90m
43 Kinematics_______________________________________________________________________________
9. A particle if projected up an inclined plane of length 20m and inclination 300 (with horizontal). What
should be the value of angle θ (with horizontal) with which the projectile be projected so that it strikes the
plane exactly at mid-point, horizontally :
A) θ = tan-1
3
2 B) θ = tan-1 (2) C) θ = tan-1 ( 3 ) D) θ = tan-1
2
3
10. A stone is projected at an angle θ with the horizontal with velocity u. It executes a nearly circular motion
near its maximum height for a short time. The radius of circular path is :
A) g2
cosu 22 θ B)
g
sinu 22 θ C)
g
cosu 22 θ D)
g
u2
11. Two projectiles A and B are fired simultaneously as shown in figure. They
collide in air at point P at time t. then :
A) t(u1 cosθ1 - u2 cosθ2) = 20
B) t(u1 sinθ1 - u2 sinθ2) = 10
C) both (A) and (B) are correct
D) both (A) and (B) are wrong
12. Two stones are projected simultaneously with equal speeds from point on an inclined plane along the line
of its greatest slope upwards and downwards repectively. The maximum distance between their points of
striking the plane is double that of when they are projected on a horizontal ground with same speed. If one
strikes the plane after two seconds of the other, the angle of inclination of plane is
A) 300 B) 450 C) 350 D) 150
13. A particle is projected under gravity with velocity ga2 from a point at a height h above the level plane.
The maximum range R on the ground is
A) h)1a( 2 +
B) ha 2
C) ha
D) 2 )ha(a +
14. The co-ordinates of a particle moving in a plane are given by x = a cos pt and y = b sin pt where a, b (< a)
and p are positive constants of appropriate dimensions. Then
A) The path of the particle is an ellipse
V= ag2 θ
44 Kinematics_______________________________________________________________________________
B) The velocity and acceleration of the particle are normal to each other at t = π/2p
C) The acceleration of the particle is always directed towards a fixed point
D) The distance traveled by the particle in time interval t = 0 to t = π/2p is a.
1 2 3 4 5 6 7 8 9 10 11 12 13 14
B BD B A D C D C A C B B D
MULTIPLE ANSWER TYPE QUESTIONS
1. A particle of mass m is thrown up vertically with velocity u. As air exerts a constant force F, the particle returns
back at the point of projection with velocity v after attaining maximum height h, then
A) )m/Fg(2
uh
2
+= B)
)m/Fg(2h
2
−=
v C)
)m/Fg(
)m/Fg(u
+
−=v D)
)m/Fg(
)m/Fg(u
−
+=v
2. A particle of mass m moves on X-axis as follows; it starts from rest at t = 0 from the point x = 0, and comes
to rest at t = 1 at the point x = 1. No other information is available about its motion at intermediate times (0
< t < 1). If α denotes the instantaneous acceleration of the particle, then
A) α cannot remain positive for all t in the interval 0 ≤ t ≤ 1
B) |α| cannot exceed 2 at any point in its path
C) |α| must be ≥ 4 at some point or points in its path
D) α must change sign during the motion, but no other assertion can be made with the information given
3. Two trains are travelling along a straight track one behind the other. The first train is travelling at 12 m/s.
The second train, approaching from the rear is travelling at speed v > 12 m/s when the second train is 200
m behind the first, the driver of second train applies brakes producing a uniform deceleration of 0.20 m/s2.
Then
A) If v = 20 m/s, the trains will not collide
B) If v = 20 /s, the trains will collide after about 20 s
C) If v = 27 m/s, the trains will not collide
D) If v = 27 m/s, the trains will collide after about 15s
4. A particle of mass m and charge q starts from rest from origin along X-axis in a region where an electric
field E = E0 - a x exists. Here E0 and a are constant and x is the distance from the starting point. Then in
the region between x = 0 to x = 2 E0/a.
A) the speed of particle first increases, then decreases
B) the particle comes to rest at x = 2 E0/a
45 Kinematics_______________________________________________________________________________
C) the particle has maximum speed at x = E0/a
D) the particle is subjected to an acceleration which changes sign at x = E0/a
5. Two particles are projected from the same point with the same speed v0 at the different angles α1 and α2
with the horizontal. Their respective times of flights are T1 and T2. If they have the same horizontal range,
and their maximum heights are H1 and H2 respectively, then
A) α1 + α2 = 900 B) H1 + H2 = v02/2g C) 1
2
1 tanT
Tα= D) 1
2
2
1 tanH
Hα=
6. A projectile thrown on a level surface attains a height h after t1 seconds and again after t2 seconds. If the
maximum height attained by the projectile is H after t seconds, then
A) t1t2 = g
h2 B) t1 + t2 =
g
H8 C) t2 - t1 =
g
)hH(8 − D) t2 - t = t - t1
7. Which of the following statements are true for a moving body?
A) If its speed changes, its velocity must change and it must have some acceleration
B) If its velocity changes, its speed must change and it must have some acceleration
C) If its velocity changes, its speed may or may not change, and it must have some acceleration
D) If its speed changes but direction of motion does not change, its velocity may remain constant
8. The figure shows the velocity (υ) of a particle plotted against time (t)
A) The particle changes its direction of motion at some point
B) The acceleration of the particle remains constant
C) The displacement of the particle is zero
D) The initial and final speeds of the particle are the same
9. A particle starts from the origin of coordinates at time t = 0 and moves in the xy plane with a constant
acceleration α in the y-direction. Its equation of motion is y = βx2. Its velocity component in the x-direction
is
A) variable B) β
α2 C)
β
α
2 D)
β
α
2
10. Two particles A and B start simultaneously from the same point and move in a horizontal plane. A has an
initial velocity u1 due east and acceleration a1 due north. B has an initial velocity u2 due north and
acceleration a2 due east
A) Their paths must intersect at some point
B) They must collide at some point
C) They will collide only if a1u1 = a2u2
D) If u1 > u2 and a1 < a2, the particles will have the same speed at some point of time
46 Kinematics_______________________________________________________________________________
11. Two particles are projected from the same point with the same speed, at different angels θ1 and θ2 to the
horizontal. They have the same horizontal range. Their times of flight are t1 and t2 respectively
A) θ1 + θ2 = 900 B) 12
1 tant
tθ= C) 2
2
1 tant
tθ= D)
2
2
1
1
sin
t
sin
t
θ=
θ
12. A large rectangular box falls vertically with an acceleration a. A toy gun fixed at A and
aimed towards C fires a particle P.
A) P will hit C if a = g B) P will hit the roof BC if a > g
C) P will hit the wall CD or the floor AD if a < g
D) May be either A, B, or C depending on the speed of projection of P
13. Two shells are filed from a cannon with speed u each, at angles of α and β respectively with the horizontal.
The time interval between the shots is T. They collide in mid air after time t from the first shot. Which of
the following conditions must be satisfied?
A) α > β B) t cos α = (t - T) cos β C) (t - T) cos α = t cos β
D) (usin α) t - 2
1gt2 = (u sin β) (t - T) -
2
1g (t - T)2
14. A particle moving with a speed v changes direction by an angle θ, without change in speed
A) The change in the magnitude of its velocity is zero
B) The change in the magnitude of its velocity is 2v sin θ/2
C) The magnitude of the change in its velocity is 2 v sin θ/2
D) The magnitude of the change is its velocity is v (1 - cos θ)
15. A particle of mass 'm' moves on the x-axis as follows:
It starts from rest at t = 0 from the point x = 0 and comes to rest at t = 1 at the point x = 1.
No other information is available about its motion at intermediate times ( 0 < t < 1)
If α denotes the instantaneous acceleration of the particle, then
A) α cannot remain positive for all t in the interval 0 ≤ t ≤ 1
B) α cannot exceed 2 at any point on its path
C) α must be ≥4 at some point or points in its path
D) α must change sign during the motion, but no other assertion can be made with the information given
16. A point moves such that its displacement as a function of time is given by x2 = t2 + 1. Its acceleration at
time t is
A) 3x
1 B)
2x
t− C)
3
2
x
t
x
1− D)
2x
t
x
1−
17. A body falls from a large height 'h' in 't' second. The time taken to cover the last metre is
A) gh
1 B)
gh2
1 C)
gt
1 D)
gt2
1
47 Kinematics_______________________________________________________________________________
18. A bead is free to slide down a smooth wire tightly stretched between the points
P1 and P2 on a vertical circle of radius R. If the bead starts from rest from P1, the
highest point on the circle and P2 lies anywhere on the circumference of the
circle. Then,
A) time taken by bead to go from P1 to P2 is dependent on position of P2 and
equals 2 g
Rcos θ
B) time taken by bead to go from P1 to P2 is independent of position of P2 and equals 2g
R
C) acceleration of bead along the wire is g cos θ
D) velocity of bead when it arrives at P2 is 2 gR cos θ
19. The position of a particle travelling along x axis is given by xt = t3 - 9t2 + 6t where xt is in cm and t is in
second. Then
A) the body comes to rest firstly at (3 - 7 ) s and then at (3 + 7 )s
B) the total displacement of the particle in travelling from the first zero of velocity to the second zero of
velocity is zero
C) the total displacement of the particle in travelling from the first zero of velocity to the second zero of
velocity is -74 cm
D) the particle reverses its velocity at (3 - 7 )s and then at (3 + 7 )s and has a negative velocity for (3 -
7 ) < t < (3 + 7 )
20. The velocity of a particle moving along a straight line increases according to the linear law v = v0 + kx,
where k is a constant. Then
A) the acceleration of the particle is k(v0 + kx)
B) the particle takes a time
0
1e
v
vlog
k
1to attain a velocity v1
C) velocity varies linearly with displacement with slope of velocity displacement curve equal to k
D) data is insufficient to arrive at a conclusion
21. A particle moves with an initial velocity v0 and retardation αv, where v is velocity at any istant t. Then
A) the particle will cover a total distance α
0v
B) the particle continues to move for a long time span
C) the particle attains a velocity 0v2
1 at t =
α
1
D) the particle comes to rest at t = α
1
48 Kinematics_______________________________________________________________________________
22. An aeroplane flies along a straight line from A to B with a speed v0 and back again with the same speed
v0. A steady wind v is blowing. If AB = l then
A) total time for the trip is 22
0
0
vv
v2
−
l, if wind blows along the line AB
B) total time for the trip is 22
0 vv
2
−
l, if wind blows perpendicular to the line AB
C) total time for the trip decreases because of the presence of wind
D) total time for the trip increases because of the presence of wind
23. At the instant a motor bike starts from rest in a given direction, a car
overtakes the motor bike, both moving in the same direction. The
speed time graphs for motor bike and car are represented by OAB and
CD respectively. Then
A) at t = 18 s the motor bike and car are 180 m apart
B) at t = 18 s the motor bike and car are 720 m apart
C) the relative distance between motor bike and car reduces to zero at t = 27 s and both are 1080 m far from
origin
D) the relative distance between motor bike and car always remains same
24. A particle having a velocity v = v0 at t = 0 is decelerated at the rate a = α v , where α is a positive
constant
A) The particle comes to rest at t = α
0v2
B) The particle will come to rest at infinity
C) The distance traveled by the particle is α
230v2
D) The distance travelled by the particle is α
230v
3
2
25. Two particles P and Q move in a straight line AB towards each other. P starts from A with velocity u1 and
an acceleration a1. Q starts from B with velocity u2 and acceleration a2. They pass each other at the
midpoint of AB and arrive at the other ends of AB with equal velocities
A) They meet at midpoint at time t = )aa(
)uu(2
21
12
−
−
B) The length of path specified i.e. AB is l = 2
21
122112
)aa(
)uaua()uu(4
−
−−
C) They reach the other ends of AB with equal velocities if (u2 + u1) (a1- a2) = 8 (a1u2 - a2u1)
D) They reach the other ends of AB with equal velocities if (u2 - u1) (a1 + a2) = 8 (a2u1 - a1u2)
49 Kinematics_______________________________________________________________________________
26. A body is moving along a straight line. Its distance xt from a point on its path at a time t after passing that
point is given by xt = 8t2 - 3t3, where xt is in metre and t is in second
A) Average speed during the interval t = 0 s to t = 4 s is 20.2 l ms-1
B) Average velocity during the interval t = 0 s to t= 4 s is -16 ms-1
C) The body starts from rest and at t = 9
16s it reverses its direction of motion at xt = 8.43 m from the start.
D) It has an acceleration of - 56 ms-2 at t = 4 s
27. Two second after projection, a projectile is traveling in a direction inclined at 300 to the horizon. After one
more second it is traveling horizontally. Then
A) the velocity of projection is 20 ms-1 B) the velocity of projection is 20 3 ms-1
C) the angle of projection is 300 with vertical
D) the angle of projection is 300 with horizon
28. A shot is fired with a velocity u at an angle (α + θ) with the horizon from the foot of an incline plane of
angle α through the point of projection. If it hits the plane horizontally then
A) tan θ = α+
α
2tan21
tan B) tan θ = 2 tan α C) tan θ =
α+
α
2tan21
tan2 D) tan θ =
α+
αα
2sin1
cossin
29. A particle is projected with a velocity 2 hg so that it just clears two walls of equal height h at horizontal
separation 2h from each other. Then the
A) angle of projection is 300 with vertical B) angle of projection is 300 with horizon
C) time of passing between the walls is g
h2
D) time of passing between the walls is 2 g
h
30. A ball starts falling freely from a height h from a point on the inclined plane
forming an angle α with the horizontal as shown. After collision with the incline it
rebounds elastically off the inclined plane. Then
A) it again strikes the incline at t = g
h8after it strikes the incline at A
B) it again strikes the incline at t = g
h2 after it strikes the incline at A
C) it again strikes the incline at a distance 4h sin α from A along the incline
D) it again strikes the incline at a distance 8h sin α from A along the incline
31. Two particles projected from the same point with same speed u at angles of projection α and β strike the
horizontal ground at the same point. If h1 and h2 are the maximum heights attained by projectiles, R be the
range for both and t1 and t2 be their time of flights respectively then
50 Kinematics_______________________________________________________________________________
A) α + β = 2
π B) R = 4 21hh C)
2
1
t
t = tan α D) tan α =
2
1
h
h
32. Two shells are fired from a cannon successively with speed u each at angles of projection α and β
respectively. If the time interval between the firing of shells is t and they collide in mid air after a time T
from the firing of the first shell. Then
A) T cos α = (T - t) cos β B) α > β
C) (T - t) cos α = t cos β
D) ( u sin α) T - 2
1gT2 = ( u sin β) (T - t) -
2
1g(T - t)2
33. Two guns situated at the top of a hill of height 10 m, fire one shot each with the same speed of 5 3 ms-1 at
some interval of time. One gun fires horizontally and other fires upwards at an angle of 600 with the
horizontal. The shots collide in mid air at the point P. Taking the origin of the coordinate system at the foot
of the hill right below the muzzle, trajectories in x - y plane and g = 10 ms-2 then
A) the first shell reaches the point P at t1 = 1 s from the start
B) the second shell reaches the point P at t2 = 2 s from the start
C) the first shell is fired 1 s after the firing of the second shell
D) they collide at P whose coordinates are given by (5 3 , 5) m
34. A radar observer on the ground is watching an approaching projectile. At a certain instant he has the
following information:
i) The projectile has reached the maximum altitude and is moving with a horizontal velocity v;
ii) The straight line distance of the observer to the projectile is l;
iii) The line of sight to the projectile is at an angle θ above the horizontal
Assuming earth to be flat and the observer lying in the plane of the projectile's trajectory then,
A) the distance between the observer and the point of impact of the projectile is
D = g
sinv2 θ - l cos θ
B) the distance between the observer and the point of impact of the projectile is D = vg
sin2 θl -
l cos θ
C) the projectile will pass over the observer's head for l < g
sectanv2 2 θθ
D) the projectile will pass over the observer's head for l > g
sectanv2 2 θθ
35. A projectile is thrown with an initial velocity u, at an angle of projection θ first from the equator and then
from the pole. The fractional decrement in the range of projectile is
51 Kinematics_______________________________________________________________________________
A) 291
1
R
dR−= B)
291
1
R
dR= C)
−=
ep g
1
g
1g
R
dR D)
−−=
ep g
1
g
1g
R
dR
36. A boat is moving directly away from a cannon on the shore with a speed v1. The cannon fires a shell with a
speed v2 at an angle α and the shell hits the boat. Then,
A) the shell hits the boat when the time equal to g
sinv2 2 α is lapsed
B) the boat travels a distance g
sinvv2 21 αfrom its original position
C) the distance of the boat from the cannon at the instant the shell is fired is
g
2(v2 sin α) (v2 cos α - v1)
D) the distance of the boat from the cannon when the shell hits the boat is
g
2(v2 sin α) (v2 cos α )
37. A projectile is fired upward with velocity v0 at an angle θ and strikes a point P (x, y) on the roof of the
building (as shown). Then,
A) the projectile hits the roof in minimum time if θ + α = 2
π
B) the projectile hits the roof in minimum time if θ + α = 4
π
C) the minimum time taken by the projectile to hit the roof is tmin = −0v
α
α−
cosg
cosgh2v 220
D) the projectile never reaches the roof for v0 < gh2 cos α.
38. A shell fired along a parabolic path explodes into two fragments of equal mass at the top of the trajectory.
One of the fragments returns to the point of firing having retraced its original path. If v is the velocity of
projectile at highest point, then
A) after explosion the other fragment has a velocity v along + x axis
B) after explosion the other fragment has velocity 3v along + x axis
C) after explosion both the fragments reach the ground with separation 2R between them, where R is the
range of the projectile
D) after explosion both the fragments hit the ground simultaneously at t = v2
R
52 Kinematics_______________________________________________________________________________
39. Identify the correct statements for a moving particle along any arbitrary path is/are:
A) avav VV =r
B) avav VV ≤r
C) avav VV ≥r
D) avav VV <r
40. The distance traveled by a moving train in the first, second and third seconds are 20m, 15m and 10m
respectively. It means that
A) the train is decelerating B) the initial velocity of the train is 20 ms-1
C) the constant deceleration of the train is 5 ms-2
D) the constant deceleration of the train is 10 ms-2
41. The displacement of a particle moving along a straight line as a function of time t is given by x = t3 + 2t2 +
3t + 4.
A) the particle starts from origin B) the initial velocity of the particle is 3 ms-1
C) the acceleration of the particle is constant
D) the particle starts with an acceleration of 4 ms-2
42. Choose the correct statement(s):
A) When the total area of the acceleration-time graph is negative, it always mean that the final velocity of
the particle is negative
B) When the total area of the velocity-time graph is negative, it always mean that the final displacement of
the particle is negative
C) When the total area of the velocity-time graph is negative, it may happen that the particle returns to its
original position
D) When the total area of the acceleration-time graph is negative, it may happen that the final velocity of
the particle is zero
43. Choose the correct statement(s) related to the v-t and x-t graphs.
A) The positions where the v-t graph interests or touches the time axis, the x-t graph shows either its
maximum or minimum
B) In the region where the v-t graph is parallel to the time axis, the particle remains stationary
C) In the region where the v-t graph coincides with the time axis, its x-t graph becomes parallel to the time
axis
D) If the v-t graph is parabolic then the acceleration is uniformly varying
44. Choose the correct statement(s) related to region A.
A) The particle starts from rest
B) The velocity of the particle is decreasing with time
C) The acceleration of the particle is negative
D) At t = T, velocity of the particle is zero
45. In the above problem choose the correct statement(s) related to region C.
A) The particle starts from rest, i.e., at t = 2T, v = 0
53 Kinematics_______________________________________________________________________________
B) The motion of the particle is accelerated
C) The particle is moving with constant velocity
D) The particle moves with increasing velocity
46. Choose the correct statement(s).
A) A projectile is a freely falling body
B) The acceleration of a projectile is maximum at the highest position
C) The time of flight is maximum when the range of the projectile is maximum
D) The normal acceleration of the projectile is maximum at its highest position
54 Kinematics_______________________________________________________________________________
COMPREHENSION TYPE QUESTIONS
Passage I (Q.No: 1 to 3):
A policeman is in pursuit of a thief. Both are running at 5
m/s. Suddenly they come across a gap between buildings
as shown in figure. The thief leaps at 5 m/s and at 450
while the policeman leaps horizontally.
1. Does the policeman clear the gap
A) No, he will not even reach at the corner of B
B) No, he will not be in a condition to cross more than 4 mts.
C) Yes, he just reaches the tip C D) Yes, he will cross the gap
2. By how much does the thief clear the gap ?
A) 0.21 m B) 0.56 m C) 0.11 m D) 0.31 m
3. By how much does the policeman fall/clear the gap ?
A) Clear by 0.09 mts. B) miss by 0.09 mts.
C) Clear by 0.90 mts. D) miss by 0.90 mts.
Passage II (Q.NO: 4 to 7):
An automobile and a truck start from rest at the same instant, with the automobile initially at some distance
behind the truck. The truck has a constant acceleration of 2.10 m/s2, and the automobile an acceleration of
3.40 m/s2. The automobile overtakes the truck after the truck has moved 40.0 m.
4. How much time does it take for the automobile to overtake the truck ?
A) 20 sec. B) 18.7 sec. C) 2 sec. D) 6.1 sec.
5. How far was the automobile behind the truck initially ?
A) 24.8 m B) 30 m C) 50 m D) 60 m
6. What is the speed of each when they are abreast ?
A) 0 & 13 m/s B) 20 m/s & 10 m/s
C) 21 m/s & 13 m/s D) 25 m/s & 30 m/s
7. Which is the best possible graph ?
55 Kinematics_______________________________________________________________________________
Passage III (Q.No: 8 to 11):
Two cars, A and B, travel in a straight line. The distance of A from the starting point is given as a function
of time by xA (t) = αt + βt2, with α = 2.60 m/s and β = 1.20 m/s2. The distance of B from the starting point
is xB (t) = γt2 - δt3, with γ = 2.80 m/s2 and δ = 0.20 m/s3.
8. Which car is ahead just after they leave the starting point ?
A) Car A moves ahead B) Car B moves ahead
C) Car A and B move simultaneously D) Data insufficient
9. At what time(s) are the cars at the same point ?
A) 2.27 sec. B) 5.73 sec. C) 2.6 sec
D) Both 2.27 sec. and 5.73 sec E) Both 5.73 sec. and 2.6 sec.
10. At what time(s) is the distance from A to B neither increasing nor decreasing ?
A) 1.66 sec. B) 1 sec. C) 4.33 sec.
D) Both 1.66 sec. and 1 sec. E) Both 1 sec. and 4.33 sec.
11. At what time(s) do A and B have the same acceleration ?
A) 6.27 sec. B) 4.33 sec. C) 2.67 sec.
D) Both 6.27 sec. and 4.33 sec. E) Both 4.33 sec. and 2.67 sec.
Passage IV (Q.No: 12 to 15):
Two bullets are shot at the same time t = 0. Their position and velocity at t = 0 are given by
,m0)0t(r1 ==→
s/mz30x50)0t(v1 +==→
mx10)0t(r2 ==→
s/mz50y40)0t(v 2 +==→
The force of gravity is working on both of them, causing an acceleration of 2s/mz10a −=r
12. )t(rand)t(r 21
→→
are
A) zt5t150yt40x10;zty5t130xt50 2−++−+ B) zt47yt62xt27;zt24x7 2+++
C) z)t5t50(yt15x10;zt5xt50 22 −++− D) None of these
13. The distance between the bullets as a function of time is
A) B) C) D)
56 Kinematics_______________________________________________________________________________
A) 3t595 2 − B) 1t10t4510 2 +− C) t3 - 7t2 + c D) 24 (t3 - 6t4 2 + )
14. At what time will be the distance between the bullets be nomial
A) 2 sec B) 1.2 sec C) 1/9 sec D) 2/7 sec
15. The relative velocity between the bullets as a function of time is
A) z20y40x50 −− B) z42x23 + C) z4x3 − D) z63y4x5 −−
Passage – V (Q.No: 16 to 20) :
An object 'P' is travelling with a velocity of 0v horizontally from
the left as shown in the figure. It is constrained to move along a
smooth track. At 'A' it can continue to travel along one of the
four paths, [ ]δγβα ,,, , as indicated in the diagram. It is given
that all the four paths, viz. γβα ,, and δ are in the same
vertical plane. All the four paths join at 'B' after which the object continuous moving horizontally. The
time taken to traverse the horizontal distance AB along the path ''α is 'T'. A detail of the four different
paths is given as:
→α is the horizontal path.
β → is a 'hill' path.
→γ is a shallow valley which can be assumed to have the same length as the horizontal path 'AB'.
→δ is a deep valley path, with a minimum depth 'H'.
Assume the total energy (E) is to mass (m) ratio of the particle as '2/'µ .
16. The ratio of the velocity of the particle at a height of 'h' on β and the velocity of the particle at a depth of
'h' on γ is:
A)
−
+
gh
gh
2
2
µ
µ B)
+
−
gh
gh
2
2
µ
µ C)
2/1
2
2
−
+
gh
gh
µ
µ D)
2/1
2
2
+
−
gh
gh
µ
µ
17. If the time taken by the object to travel from A to B along β is βt , then:
A) Ttt == αβ B) Ttt => αβ C) Ttt =< αβ D) cannot be predicted
18. If the time taken by the object to travel from A to B along γ is γt , then :
A) Ttt == αγ B) Ttt => αγ C) Ttt =< αγ D) cannot be predicted
19. If the time taken by the object to travel from A to B along δ is δt , then :
57 Kinematics_______________________________________________________________________________
A) Ttt == αδ B) Ttt => αδ C) Ttt =< αδ D) cannot be predicted
20. If we approximate the time taken to travel from A to B along the deep path ''δ by the time taken for a
vertical return path of depth 'H', i.e., from A and back to A, by assuming the initial velocity to be negligible
as compared with that for the major portion of the path, then the total time taken is:
A) g
H2 B)
g
H4 C)
g
H8 D) T
Passage VI (Q.No: 21 to 23):
A point moves rectilinearly in one direction. Figure shows the distance s
traversed by the point as a function of time t. Using this graph, answer the
following questions.
21. The average velocity of the point during the time of motion is
A) 10 cm/s B) 15 cm/s
C) 20 cm/s D) 25 cm/s
22. The maximum velocity is
A) 15 cm/s B) 20 cm/s
C) 25 cm/s D) 30 cm/s
23. The time moment 0t at which the instantaneous velocity is equal to the mean velocity averaged over the
first 0t second is
A) 10s B) 16s C) 18s D) 20s
Passage VII (Q.NO 24 to 26):
Two particles are initially located at points A and B a distance d
apart as show in figure. They start moving at time t = 0 such that
the velocity u of B is always along the horizontal direction and
velocity v of A is continually aimed at B as shown in figure.At
time t = 0, u is perpendicular to v.
24. About the velocities u and v, we can say that
A) both the velocities are constant
B) both the velocities are changing
C) the velocity u is constant while the velocity v is changing
D) the velocity v is constant while the velocity u is changing
25. The relative velocity of approach of A towards B is
58 Kinematics_______________________________________________________________________________
A) (v - u cos θ ) B) (v + u cosθ ) C) (u - v cosθ ) D) (u + v cos θ )
26. The particles A and B will meet after a time
A) 22 uv
vd
+ B)
( )22 uv
vd
− C)
( )vd
uv 22 + D)
( )vd
uv 22 −
Passage - VIII (Q.No: 27 to 29) :
Figure shows the velocity time graph of a particle moving along a straight line.
Answer the following questions.
27. The region in which the rate of change of velocity of the
particle is maximum.
A) 0 to 2 s B) 2 to 4 s
C) 4 to 6 s D) 6 to 8 s
28. The particle comes to rest at time
A) 0 sec B) 4.67 sec
C) 5 sec D) 8 sec
29. The maximum displacement of particle is
A) 33.3 m B) 20.2 m C) 26.6 m D) zero
PASSAGE - IX (Q.No: 30 to 33):
On the bank of a river two swimmers made a challenge as "who
will reach the point B on the other bank early?" So both of them
start from point 'A' on one bank of the river to reach the winning
point B on the other bank, lying directly opposite to point A. The
stream velocity was known to be 2 km/hr and the velocity of both
the swimmers in still water was 2.5 km/hr. Both of them follow
different paths to reach point B the swimmer 'S1' crosses the river along straight line AB, while the other
swimmer 'S2' swims at rights to the stream and then walks the distance which he has been carried away by
the stream to go to the point 'B'. Assume the velocity (uniform) of his walking as (2/3) km/hr and the width
of the river as 'W'.
30. The value of angle ''θ shown in the figure is:
A) cos-1 (4/5) B) cos-1 (3/4) C) sin-1(4/5) D) sin-1(3/4)
31. The value of angle ''φ shown in the figure is :
59 Kinematics_______________________________________________________________________________
A) sin-1 (4/5) B) cos-1 (4/5) C) tan-1 (4/5) D) data insufficient
32. Match column I with column II in reference to the passage.
Column I Column II
I Time for S1 to reach B (A)
5
4W
II Time for S2 to reach C (B)
5
6W
III Drift 'x' for S2 (C)
3
2W
IV Time for S2 to reach B
from C
(D)
15
8W
V Difference in time, t∆ , for
S1 and S2 to reach B
(E)
5
2W
A) I-C, II-E, III-A, IV-B,V-D B) I-A, II-B, III-D,IV-C, V-E
C) I-E, II-C, III-B, IV-D, V-A D) I-B, II-A, III-D, IV-C, V-E
33. What should be the velocity (assume uniform) of walking of swimmer 'S2' such that both the swimmers
reach at the point B simultaneously?
A) 1.2 km/hr B) 2.7 km/hr C) 3 km/hr D) needs more information
* * *
MULTIPLE MATCHING TYPE QUESTIONS:
1. Match the following:
List - I List - II
a) Range = height (Max) e) acceleration perpendicular to velocity
b) For θ = 450, at the highest point f) R =
g17
8
c) y = px + qx2 g) Kmin =
2
1 (Kmax)
d) At t = T/2 (for projectile) h) θ = Tan-1 (4)
i)
1Pg
P2T2 +
=u
2. Match the following:
60 Kinematics_______________________________________________________________________________
List - I List - II
a) one dim. motion e) motion of one projectile w.r.t another projectile
b) for no air resistance f) time of ascent = Time of decent for a vertically projected up body
c) for projectile H = R/4 g) T = 2u/g, for a body vertically projected up
d) Range = g
u2
i) θ = 450
3. Match the following:
List - I List - II
a) v - t graph e) Area gives displacement
b) x - t - graph f) slope gives velocity
c) a - t graph g) slope gives acceleration
d) a α x h) Area gives change in velocity
h) slope of v - x graph is constant
t) non uniform acceleration
4. Match the following:
List - I List - II
a) Magnitude of acceleration is constant e) uniform circular motion
b) tangential acceleration is zero f) non-uniform circular motion
c) speed is constant g) projectile motion
d) angle between radial acceleration and velocity is 900 h) accelerated straight line motion
* * *
61 Kinematics_______________________________________________________________________________
KEY
Multiple Answer Type Questions:
1. ABC
2. AC
3. ABCD
4. ABCD
5. ABCD
6. ABCD
7. AC
8. ABCD
9. D
10. ACD
11. ABD
12. ABC
13. ABD
14. AC
15. AC
16. AC
17. BC
18. BCD
19. ACD
20. ABC
21. AB
22. ABD
23. AC
24. AD
25. ABC
26. ABCD
27. BC
28. AD
29. AD
30. AD
31. ABCD
32. ABD
33. ABCD
34. BC
35. BD
36. ABCD
37. ACD
38. BCD
39. AC
40. AC
41. BCD
42. CD
43. ACD
44. BCD
45. ABD
46. AD
Comprehension Type Questions:
1. B
2. D
3. B
4. D
5. A
6. C
7. A
8. A
9. D
10. E
11. C
12.
13.
14.
15.
16.
17.
18.
19.
20.
21. A
22. C
23. B
24. C
25. A
26. B
27. C
28. A,B,D
29. A
30.
31.
32.
33.
Multiple Matching Type Questions:
1. a → f, h
b → e, g
c → i
d → e
2. a → e, h, g, f
b → f, g
c → i
d → i
3. a → e, g
b → f
c → h
d → i, j
4. a → e, g, h
b → e
c → e
d → e, f
![KINEMATICS - new.excellencia.co.innew.excellencia.co.in/college/web/pdf/Kinematics-merged.pdf · KINEMATICS KINEMATICS WORKSHEET 1 1) Displacement is a _____ [ ] 1) Vector quantity](https://img.pdfslide.us/doc/110x75/5f356d4687229051801abace/kinematics-new-kinematics-kinematics-worksheet-1-1-displacement-is-a-.jpg)
