3 Kinematics

Syllabus: Motion in a straight line, uniform motion, its graphical representation, uniform accelerated motion and its applications,

projectile motion.

Review of Concepts

1. Time : It is measure of succession of events. It is a

scalar quantity. If any event is started at t - 0 then time will

not be negative. But if the oDservation is started after the start

of event then time m a y be negative.

2. Distance and Displacement: Suppose an insect is at

a point A (Xj, t/1; Zj) at t = tp It reaches at point B (x2, yi, Z2) at

t = f2 through path ACB with respect to the frame shown in

figure. The actual length of curved path ACB is the distance

travelled by the insect in time At = t2 - fj.

C

> - X

If we connect point A (initial position) and point B (final

position) by a straight line, then the length of straight line

AB gives the magnitude of displacement of insect in time

interval At = t 2 - tp

The direction of displacement is directed from A to B

through the straight line AB. From the concept of vector, the

position vector of A is = x 1 t + 2/if+Z;[ it and that of B is

—> A A A rB = * 2 l + l/2j+Z2k.

According to addition law of vectors,

rA + AB = rB

— » — >

AB = rg - rA

= (X2~X1) l + (t/2-yi) ) + (Z2-Zl) k

The magnitude of displacement is

IABI = V ( x 2 - x 1 ) 2 + ( y 2 - y i ) 2 + ( 2 2 - 2 i ) 2

Some Conceptual Points :

(i) Distance is a scalar quantity.

(ii) Distance never be negative.

(iii) For moving body, distance is always greater than zero.

(iv) Distance never be equal to displacement.

(v) Displacement is a vector quantity.

(vi) If a body is moving continuously in a given

direction on a straight line, then the magnitude of

displacement is equal to distance.

(vii) Generally, the magnitude of displacement is less or

equal to distance.

(viii) Many paths are possible between two points. For

different paths between two points, distances are

different but magnitudes of displacement are same.

The slope of distance-time graph is always greater

or equal to zero.

The slope of displacement-time graph may be

negative.

Example : A man walks 3 m in east direction, then 4 m

north direction. Find distance covered and the

displacement covered by man.

Solution: The distance covered by man is the length of

path = 3 m + 4 m = 7m.

N

(ix)

(x)

in

Let the man starts from O and reaches finally at B (shown

in figure). OB represents the displacement of man. From

figure,

I OB I = {OA)2 + (AB)2

= (3 m)2 + (4 m)2 = 5 m

and tan 6 = 4 m 4

3 m 3

) = tan' - 1 3

v /

The displacement is directed at an angle tan 1 T 3

v y

north

of east.

3. Average Speed and Average Velocity : Suppose we

wish to calculate the average speed and average velocity of

the insect (in section 2) between i - 1 ] and t = t2. From the

path (shown in figure) we see that at t = f j , the position of

Kinematics 35

the insect is A (xa, y\, zx) and at t = t2, the position of the insect

is B (x2, y2, z2).

The average speed is defined as total distance travelled

by a body in a particular time interval divided by the time

interval. Thus, the average speed of the insect is

The length of curve ACB vav = : 7

t2 - h

The average velocity is defined as total displacement

travelled by a body in a particular time interval divided by

the time interval.

Thus, the average velocity of the insect in the time

interval t2 - tj is — »

-4 AB

V a V ~ ' 2 - t 2

—» —¥ =

rB~ rA

- t2—t]

= (*2 - * l ) 1 + (yi - ?/l) ) + (22 - Zl) ^

h~h

Some Important Points:

(i) Velocity is a vector quantity while speed is a scalar

quantity.

(ii) If a particle travels equal distances at speeds

i>i, v2, v3, ... etc. respectively, then the average

speed is harmonic mean of individual speeds.

(iii) If a particle moves a distance at speed V\ and comes

2v\v2 back with speed v2, then vav = -

Mathematically,

Vi +v 2

But Vav = 0 (iv) If a particle moves in two equal intervals of time at

different speeds v^ and v2 respectively, then

Vi + v2

(v) The average velocity between two points in a time

interval can be obtained from a position versus

time graph by calculating the slope of the straight

line joining the co-ordinates of the two points.

(t2-tn) : ( X 2 - X l )

(b)

The graph [shown in fig. (a)], describes the motion

of a particle moving along x-axis (along a straight

line).

Suppose we wish to calculate the average velocity

between t = t j and t = t2. The slope of chord AB

[shown in fig. (b)] gives the average velocity.

: tan 0 = t2-h

(vi) The area of speed-time graph gives distance.

(vii) The area of velocity-time graph gives displacement.

(viii) Speed can never be negative.

4. Instantaneous Velocity : Instantaneous velocity is

defined as the average velocity over smaller and smaller

interval of time.

Suppose position of a particle at t is ~r*and at t + At is

r + A r. The average velocity of the particle for time interval

. . . -> Ar S V f l t , = AT

From our definition of instantaneous velocity, At should

be smaller and smaller. Thus, instantaneous velocity is

-» .. Ar dr V = H M AT = Tt

If position

-xi + y^ + zi

of

A A lie, then

A f - > 0

a particle

x-component of velocity is vx = ~

y-component of velocity is vy =

z-component of velocity is vz =

Thus, velocity of the particle is

at an instant t

dx

is

—> A A V = VXl + VY)+VZ,

dx A dy

dt

A d z A 1 + di%

Some Important Points:

(i) Average velocity may or may not be equal to

instantaneous velocity.

If body moves with constant velocity, the

instantaneous velocity is equal to average velocity.

The instantaneous speed is equal to modulus of

instantaneous velocity.

Distance travelled by particle is

s = J \v\ dt

x-component of displacement is Ax = J vx dt

y-component of displacement is Ay = J Vy dt

z-component of displacement is Az = jvzdt

Thus, displacement of particle is

A~r = Axt + Ayf + Az 1c

If particle moves on a straight line, (along x-axis),

dx

(ii)

(iii)

(iv)

(v)

(vi)

then v = dt

(vii) The area of velocity-time graph gives displacement.

(viii) The area of speed-time graph gives distance.

(ix) The slope of tangent at position-time graph at a

particular instant gives instantaneous velocity at

that instant.

5. Average Acceleration and Instantaneous Acceleration :

In general, when a body is moving, its velocity is not always

36 Kinematics

the same. A body whose velocity is increasing is said to be

accelerated.

Average acceleration is defined as change in velocity

divided by the time interval.

Let us consider the motion of a particle. Suppose that the

particle has velocity "vj at f = fi and at a later time t = t2 it has

velocity ~V2- Thus, the average acceleration during time

interval At = t2 - t\ is

V 2 ~ V 1 A V

a " v ~ h - h ~ At

6. Problem Solving Strategy :

If the time interval approaches to zero, average

acceleration is known as instantaneous acceleration.

Mathematically,

,. Av dv a= lim —— = "37

AT->OA' DT

Some Important Points:

(i) Acceleration a vector quantity.

(ii) Its unit is m/s .

(iii) The slope of velocity-time graph gives acceleration.

(iv) The area of acceleration-time graph in a particular

time interval gives change in velocity in that time

interval.

Motion on a Straight Line (one dimensional motion)

Uniform velocity

(i) s = vt

(ii) a = 0

Motion with constant

acceleration f

(i) s = u + v

(i) If «=/(*) ,« =

Motion with variable

acceleration

dv

dt

1 7 (ii) s = ut + ^ ar

(iii) v2 = u2 + las

(iv) v = u+at

( v ) S „ t h = M + ( 2 « - l ) |

(vi) For retardation, 'a' will be

negative.

7. Motion in Two or Three Dimension : A body is free

to move in space. In this case, the initial position of body is

taken as origin.

Any convenient co-ordinate system is chosen. Let us

suppose that at an instant t, the body is at point P (x, y, z).

The position vector of the body is r = x t + + z it. Thus,

velocity

dr dx A A dz

(ii) If a =/(s), a = v dv

(iii) If a =/&>),« =

ds

dv

dt

(iv) v = ds

dt

(v) s = { vdt

(vi) v-jadt

Discussion:

(i) (a) If ax is constant,

vx = ux + axt

1 .2 x = uxt + -axt

vx=ux + laxx

In this way,

„ A az A

dx

2 _ . . 2 x'

(b) If ax is variable,

x = jvxdt

V* = -dt jdvx = jax dt

and acceleration along x-axis is ax =

dy The velocity along y-axis is

dvx

dt

(ii) (a) If ay is constant,

UVy

and the acceleration along y-axis is ay - — •

dv.

dv^

dt

1 x2 y = uyt + -ay t

Vy = Uy+ayt

v2 = u2y + layy

(b) If ay is variable,

vz = - and a z = - d t Similarly,

The acceleration of the body is ~a = ax^ + aA + azic.

y = jvydt

jdvy = jaydt

Kinematics

(iii) (a) If az is constant,

z = u 2 f + - a 2 r

v\ = u\ + 2az z

(b) If az is variable,

z = jvzdt

jdvz = jazdt

If the motion of the body takes place in x-y plane, then

az = 0, vz = 0, uz = 0

Example: A car moves in the x-y plane with

acceleration (3 m/s2 t + 4 m/s2 f )

(a) Assuming that the car is at rest at the origin at

/ = 0, derive expressions for the velocity and position

vectors as function of time.

(b) Find the equation of path of car.

Solution : Here, ux = 0, uy = 0, uz = 0

37

(a)

or

and

or

(ii)

or

and

or

ax = 3 m/s2, ay = 4 m/s2

vx = 31

VY = UY+ayt

VY = 4t

—» A A V = Vxl+VY)

= (3fi + 4 f j )

1 a x = uxt + ^axt

*=lx3f2 = |f2

t 1 l 2 y = uyt + — ayt

y = ~(4)t2 = 2t2

o 2 4

4 = 3 *

Hence, the path is straight line,

(b) The position of car is

r = xi+yj =-ri + 2r j

Example: A bird flies in the x-y plane with a velocity

v'=/2t + 3 f A t f =0, bird is at origin.

Calculate position and acceleration of bird as function of

time.

Solution:

Here,

— >

vx = t , vu = 3t and vz = 0

or

or

or

dx 2

dt~

J M t2dt

Also,

or

or

x=-

VY = 3t

dy I'3'

3 tdt

3t y = T

.'. Position of particle is x t + y|

f 3 « 3t2 A

3 2

flr = d (t2)

dt dt = 21

and

or

VY = 3t

dvv „ dt dt dt

fly = 3 unit

— > a A

a = flxi + flyj

= 2ft + 3 j

8. Motion Under Gravity: The most familiar example

of motion with constant acceleration on a straight line is

motion in a vertical direction near the surface of earth. If air

resistance is neglected, the acceleration of such type of

particle is gravitational acceleration which is nearly constant

for a height negligible with respect to the radius of earth. The

magnitude of gravitational acceleration near surface of earth

is g = 9.8 m/s2 = 32 ft/s2.

Discussion:

Case I : If particle is moving upwards :

In this case applicable kinematics relations are:

v = u-gt (i)

(ii)

(iii)

(iv)

h = ut-±gt2

v2 = u2- 2gh

Here h is the vertical height of the particle in

upward direction. For maximum height attained by

projectile

h=K v = 0

i.e., (0)2 = u2 - 2ghmax

i —

"mav — <•» U_

2g-Case II : If particle is moving vertically downwards :

In this case,

38 Kinematics

(i)

(ii)

(iii)

v = u+ gt

v2 = u2 + 2gh

h = ut + ^gt2

:: I1

Here, h is the vertical height of particle in

downward direction.

9. Projectile Motion: A familiar example of two

dimensional motion is projectile motion. If a stone is thrown

from ground obliquily, it moves under the force of gravity

(in the absence of air resistance) near the surface of earth.

Such type of motion is known as projectile motion. We refer

to such object as projectile. To analyse this type of motion,

we will start with its acceleration. The motion of stone is

under gravitational acceleration which is constant in

magnitude as well as in direction.

Now let us consider a projectile launched so that its

initial velocity Vq makes an angle 0 with the horizontal

(shown in figure )• For discussion of motion, we take origin

at the point of projection. Horizontal direction as x-axis and

vertical direction as y-axis is taken.

The initial velocity of projectile along x-axis is

ux = Vq COS 0.

The component of gravitational acceleration along x-axis

is ax=gcos 90° = 0.

The component of initial velocity along y-axis is

Uy = Vq sin 0.

The acceleration along y-axis is Uy = -g.

Discussion:

(i) The instantaneous velocity of the projectile as

function of time : Let projectile reaches at point

(x, y) after time t (shown in figure).

and

Vx = Ux = Vq COS 0

Vy = Uy-gt = v0smQ-gt

v = vxi+vy)

~V = t>0 COS 0*1 + (UQ sin 0 - gt) f

The instantaneous speed

= l~vl = V(u0 cos 0)2 + (Vq sin 0 - gt)2

Also, X = Ux t — (Vq COS 0) t = Vqt COS 0

1 y = uyt--gt

1 2 y = (z>0sin Q)t--gt

The position of the projectile is

—> A A r = x i + y j

: Vqt COS 0 i + y o t s i n 0 - - £ (

(ii) Trajectory of projectile: The y-x graph gives the

path or trajectory of the projectile.

From discussion of instantaneous velocity of projectile.

1 2 x = uo<cos0 ...(1) and y = v0t sin 0 - - gt ...(2)

t--Vq COS 0

Putting the value of t from (3) in the equation (2),

' - ^ r x * ;g

...(3)

y = v o sin 0

or y = x tan 0 -

VQ COS 0

S * 2

Vq COS 0

2VQ COS2 0

This is the required path of projectile.

2z;2 cos2 0 Multiplying the equation (4) by

..(4)

to both sides,

we get

2 2v sin 0 cos 0 x x =

g

2v cos 0 g

Adding Vq sin 0 cos 0

g to both sides, we get

x -sin 0 cos 0

g

This is of the form,

2vq COS 2 0

y—

1 9 Vq sin 0

(x-a) =- c(y-b)

which is the equation of a parabola. Hence, the equation

of the path of the projectile is a parabola.

(iii) Time of flight: The time taken by projectile to

reach at point A from point O is known as time of

flight.

Here, OA = vx T, where T is time of flight. The total

displacement along y-axis during motion of

projectile from O to A is zero so, y = 0,

But

or

1 2

y = u y T ~ 2 s T

0 = (vq sin Q)T-jgT2

T = 2v0 sin 0

g

Kinematics 39

(iv) Range of projectile: Distance OA is known as

range.

The time taken to reach to point A from point 0 is

2I>Q sin 0 T-

The range R-uxT= (v0 cos 0) r2 vr, sin 0 '

(2 sin 0 cos 9) = vo

vq sin 20

g

The time taken by projectile to reach from O to B is equal

T to the time taken by projectile to reach from B to A = —•

(v) Height attained by projectile : At the maximum

height (at point B) the vertical component of

velocity is zero.

y

J i vy = °

B r ii B r ii

H

v2 = u2-2gH

(0)2 = (v sin 0)2 - 2gH

H =

2 2 VQ sin 0

2g Alternative method: A particle is projected with a

velocity u at an angle a to the horizontal, there being no force

except gravity, which remains constant throughout its

motion. —> A . A u =u cos a i + u sin a j y

A

—> A A s = x i + y j

— » s=l}t + \^t2 (0,0)

A A , A A. , 1 ,2A xi + y j = (u cos a i + u sin a j ) t-^g* J

x = ut cosa, y = ut sin a - ^ gt2

For the maximum height,

T t = -

rr 2u sin a R TT T = —' x = —' y = H

g 2 *

H = 2 • 2 u sin a

2^

(a) For the range,

x = R, y = 0, t = T

R = u2 sin 2a

(b) (i) If for the two angles of projection a j and a2 , the

speeds are same then ranges will be same. The

condition is a j + a 2 = 90°.

(ii) If particles be projected from the same point in

the same plane so as to describe equal

parabolas, the vertices of their paths lie on a

parabola.

(iii) The locus of the foci of all parabolas described

by the particles projected simultaneously from

the same point with equal velocity but in

different directions is a circle.

(iv) The velocity acquired by a particle at any point

of its path is the same as acquired by a particle

in falling freely from the directrix to that point.

(v) A projectile will have maximum range when it

is projected at an angle of 45° to the horizontal

and the maximum range will be

Rn

g

At the maximum range, H = -

(vi) In the case of projectile motion, at the highest

• point, potential energy is maximum and is

1 2 2 equal to — mu sin a.

(vii) If the body is projected at an angle of 45° to the

horizontal, at the highest point half of its

mechanical energy is potential energy and rest

is kinetic energy.

(viii) The weight of a body in projectile motion is

zero as it is freely falling body.

(ix) If two projectiles A and B are projected under

gravity, then the path of projectile A with

respect to the projectile B is a straight line.

10. The equation of trajectory of projectile is

X - -u sin a cos a

g

f 2 9 \ 2u cos a

g y—

2 • 2 N

u sin a

2*

(a) Latus rectum = 2 u2 cos2 a

S be the focus, then

AS = ^ (latus rectum)

SM = -2 2 u cos a

z

A

>S N. A a

2g M

(b) The co-ordinates of the focus

2 • 2 - 2 u sin a cos a —u sin a

g 2g

(c) The equation of directrix, y = —

The range of Mth trajectory

<u

40 Kinematics

e" 1 u2 sin 2a

8 where e is the coefficient of restitution.

11. Projectile motion on an inclined plane: A

projectile is projected up the inclined plane from the point

O with an initial velocity v0 at an angle 9 with horizontal.

The angle of inclination of the plane with horizontal is a (as

shown in figure)

g sin ( 9 0 ° - a )

= g cos a

l » = u t + ^ t 2

or fit+ o f = (« cos (a - P)*+ u sin (a -P) " j ) T

- | s T 2 ( s i n Pi + c o s P j )

Equating the coefficients of 1 and f separately.

We get, R = uT cos (a - P ) - 1 gT2 sin (3

1 2 0 = uT sin a - - gT cos P

. T = 2 M s i n ( a - p )

g COS P

(b) Range is R = 2u cos a sin (a + P)

g c o s 2 p

The acceleration along x-axis is U y S

ax = - g sin a and <h/ = - g cos a.

The component of velocity along x-axis is y \

(a) Time of flight: During motion from point O to A,

the displacement along y-axis is zero.

y = 0 at t = T

. 1 ,2 y=Uyt+ — Clyt

or 1 ?

0 = v0 sin (0 - a) T - ^ g cos a T z

T = 2v0 sin (0 - a )

g c o s a

(b) Range of projectile: As shown in figure OA

represents the range of projectile. For range,

x = R, t = T

1

or

x=uxt+^axt

1 ? R=v0 COS (9 - a) T - ^gsin ol T . . . (1)

2v0 sin (0 - a ) Putting the value of T = ^ ^ — » in eq. (1)

g cos a

R = -vl

[sin (20 - a) - sin a] g cos ' a

Alternative method:

Here, P = The angle of inclination of the inclined plane

a = The angle of projection

u = The velocity of projection

/. In vector form,

! i = - g sin p i - g cos p j1

u cos {d - P) 1 + u sin (d - p) j

For the point A,

t = T= the time of flight.

: — 1 [ s i n ( 2 a - p ) - s i n p] gcos P

(c) For maximum range

2 « - P = f

u2 (1 - sin p) ^max —

(d)

(e) (i)

(ii)

g cos2 p S (! + sin P)

T2g = 2Rmax

When the range of a projectile on an inclined

plane is maximum, the focus of the path is on

the plane.

From a point on the ground at distance x from

a vertical wall, a ball is thrown at an angle

45°, it just clears the wall and strikes the ground

at a distance y on the other side. Then the

xy height of the wall is •

6 x + y (iii) If a body moves along a straight line by an

2/o engine delivering constant power, then t ~ s

(iv) If a, b, c be distances moved by a particle

travelling with uniform acceleration during xth,

yth and zth second of its motion respectively,

then a (y - z) + b(z - x) + c (x - y) = 0

12. Relative velocity:

~vAB = relative velocity of A with respect to 8

VAB = - V b

Kinematics 41 — >

V B / 1 = v B - v ^

a A B = ~ a B

(a) If a satellite is moving in equatorial plane with

velocity "v and a point on the surface of earth with

velocity u relative to the centre of earth, the velocity

of satellite relative to the surface of earth

V S E : : V - U

(b) If a car is moving at equator on the earth's surface

with a velocity relative to earth's surface and a

point on the surface of earth with velocity V£ relative

to its centre, then

V C E = v c - uE

(c) If the car moves from west to east (the direction of

motion of earth)

VC = VCE + V£

and if the car moves from east to west (opposite to

the motion of earth)

»C = - »E

(d) For crossing the river in shortest time, the boat

should sail perpendicular

to the flow.

If the width of river is d.

v = the velocity of boat in

still water, then,

t J -i'

B vT C

V ' / The position of boat at the

other bank is C (not B). - » - » - >

The displacement of the boat = OC = OB + BC

OC = V(OB)2 + (BC)2

= Vi2 + (vrt)2 = Vd2 vr-V

(e) For crossing the river in shortest distance, the boat

moves as such its horizontal component of velocity

balances the speed of flow.

OB = the shortest path = d

vr = v sin 0 sin 0 = -

cos 0 = V1 - sin2 0

V -v.

t =

t =

V COS 0 v Vt;2 - £

In this case, the magnitude of displacement = d.

(f) If boat crosses the river along the shortest path, then

time is not least.

(g) If c is a space curve defined by the function r (f), then

d t .

dt is a vector in the direction of the tangent to c. If

the scalar t is taken arc length s measured from some

d~t

fixed point on c, then is a unit tangent vector to

c j m d the arc length is denoted by R. Then

= k~ii where is a unit normal vector.

ds

(h) The derivative of vector of constant magnitude is

perpendicular to the vector itself.

(i) The derivative of a vector of constant direction is

parallel to that vector.

(j) y-x curve gives actual path of the particle. The

tangent at a point o n y - x curve gives the direction

of instantaneous velocity at that point,

(k) When n number of particles are located at the

vertices of a regular polygon of n-sides having side

length a and if they start moving heading to each

other,

they must collide at the centre of polygon after the

where v is speed of each

time t = -1 271 1 - COS

particle.

13. Velocity of approach: If two particles A and B

separated by a distance d at a

certain instant of time move v2

with velocities Vi and v2 at /

angles 0j and 02 with the b ~

direction AB, the velocity by

which the particle A approaches B = Uj cos 0! - v2 cos 02.

The angular velocity of B with respect to A

z>2 sin 02 - sin 0j = _

Example : Four particles are located at the corners of a

square whose side equals a. They all start moving

simultaneously with velocity v constant in magnitude, with

the first particle heading continually tor the second, the

second for the third, third for the fourth and fourth for the

first. How soon will the particles converge?