Particle Dynamics 1 C O N S E R V A T IO N L A W...Target Course MASTERS ACADEMY 1-Anasagar Circular Road, Opp. Chaupati, Ajmer. 0145-2633111 Particle Dynamics 3 C O N S E R V A T

MASTERS ACADEMY 1-Anasagar Circular Road, Opp. Chaupati, Ajmer. 0145-2633111Target Course

C O N S E R V A T I O N L A W1Particle Dynamics

Introduction

The branch of physics which deals with the study of motion of objects is called Mechanics.

Branches of mechanics - Mechanics

Statics Dynamics

Kinematics KineticsIt is branch of mechanicswhich is concerned aboutthe causes (i.e. the force ,torque) that causemotion of bodies.

The word kinematics means‘science of motion’. Branch ofmechanics which deals withstudy of motion without goinginto the cause of motion, i.e.force, torque etc.

Study of forces and theireffect on objects at rest

Study of forces and theireffect on objects in motion

Rest and Motion

Rest An object is said to be at rest if it does not change its position with time, with respect toits surroundings.

t1 t2 t3

x

y

S

Motion An object is said to be in motion if it changes its position with time, with respect to itssurroundings.

A

O Time

Whensittingona chair, your

speed is zero relative to Earth

but 30 km/s relative to the Sun.

Rest and Motion are relative that means An object in one situation can be at rest but in anothersituation the same object can be in motion. e.g. a person sitting in a moving train is at rest withrespect to his fellow passengers but is in motion with respect to the objects outside the train.

e.g.



Types of Motion in a Body(1) Rectilinear or Translatory Motion

Rectilinear Motion: Motion in which a particle or point mass body is moving along a straight line.

Translatory Motion: Motion in which each particle of the body has the same displacement in thesame time interval.

(2) Circulatory or Rotatory Motion

Circulatory Motion: Motion in which a particle or point mass body is moving on a circle.

Rotatory Motion: Motion in which each particle of the body move in a circle.

(3) Oscillatory or Vibratory Motion

Oscillatory Motion: Motion in which a body moves to and fro or back and forth repeatedly about afixed point.

Vibratory Motion: Oscillatory motion of body in which the amplitude is very small.

Point Mass ObjectWhen an object in motion covers a very large distance as compared to its size, the object can beregarded as a point object.

e.g. A car travelling a few hundred kilometre distance, can be taken as a point object. But if a caris travelling a distance which is not very large as compared to the size of the car, then the car cannot be taken as a point object.

Frame of ReferenceThe place from which motion is observed and measured is called frame of reference.

The Frame of reference is a system of coordinate axes attached to an observer from which,theobserver can describe position, displacement, acceleration etc.of a moving object.

Position

The position of a particle refers to its location in the space at a certain moment of

time. It is concerned with the question _ “where is the particle at a particular moment of

time?”

Motion in One, Two and Three Dimensions

One Dimensional Motion Two Dimensional Motion Three Dimensional Motion

Only one of the three coordinates Two of the three coordinates All the three coordinates

changes with respect to time. change with respect to time. change with respect to time.

e.g.(i) A train running on a straight e.g.(i) A point moving in a plane e.g.(i) Objects moving in space

railway track (ii) The Earth revolving around (ii) A kite flying on a windy day

(ii) A bus moving straight the Sun.

on a level road.



Distance and Displacement

(i) Distance The total length of the path travelled by an object during a given time is called thedistance.

It is a scalar quantity.

SI unit : metre

The value of distance cannot be negative.

(ii) Displacement The shortest distance from the initial position to the final position of an objectis called displacement.

It is a vector quantity i.e. it has magnitude as well as direction.

SI unit : metres

The value of displacement can be zero, positive or negative.

Difference Between Distance and DisplacementDistance Displacement

1. Total path travelled by the object in given time. 1. Shortest distance from initial to final position of the object in given time.

2. It is a Scalar quantity. 2. It is a Vector quantity.

3. It is never negative or zero, but is always positive. 3. It can be positive, zero or negative.

4. The distance is either equal or greater than 4. The displacement can be equal or or less than

displacement but never less than displacement. distance travelled but never greater than

distance travelled.

5. The distance travelled by the object between two 5. The displacement of an object between twoposition tells the type of path followed. positions has a unique value and does not

tell the type of path followed.

6. Distance never decreases with time. 6. Displacement decreases when the particle is moving towards the initial point.

Check Point 1:- Ram takes path 1 (straight line) to go from P to Q and Shyam takes path2 (semicircle).(a) Find the distance travelled by Ram and Shyam?

P Q1

2

100 m

(b) Find the displacement of Ram and Shyam?Sol: (a ) Distance travelled by Ram = 100 m

Distance travelled by Shyam = (50 m) = 50 m(b) Displacement of Ram = 100 m

Displacement of Shyam = 100 m

Check Point 2:- An old person moves on a semi circular track of radius 40 m during a morning walk.If he starts at one end of the track and reaches at the other end. Find the displacement of theperson.

Sol. Displacement = 2R = 2 × 40 = 80 meter.

Check Point 3:- An athelete is running on a circular track of radius 50 meter. Calculate the displacementof the athlete after completing 5 rounds of the track.

Sol. Since final and initial positions are same.

Hence displacement of athlete will be r r r 0 Check Point 4:- If a particle moves from point A to B then distance covered by particle will be.



Sol. D = x + 2x = 3x

Check Point 5:- A monkey is moving on circular path of radius 80 m. Calculate the distancecovered by the monkey in a complete cycle.

Sol. Distance = Circumference of the circle

D 2 R D 2 80 = 160 × 3.14 = 502.40 m

Check Point 6 :- A man has to go 50 m due north, 40 m due east and 20 m due south to reach a field.

(A) What distance he has to walk to reach the field?

(B) What is his displacement from his house to the field?

Sol. Let origin be O then

(A) Distance covered = OA + AB + BC

= 50 + 40 + 20

= 110 meter

(B) First method : Second method :

Displacement 2 2OC OD CO Displacement ˆ ˆ ˆd 50j 40i 20 j

2 240 30 ˆ ˆ30j 40i

10 25 50 meter 2 2| d | 40 30 50

meter

Check Point 7:- A body covers1

4th part of a circular path. Calculate the ratio of distance and

displacement.

Sol. Distance = AB from path (1) Displacement = AB

2 r r

4 2

2 2OA OB

2 2r r r 2

Dis tan ce r / 2

Displacement r 2 2 2

Check Point 8:- A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward,followed again by 5 steps forward and 3 steps backward and so on. Each step is 1 m long andrequires 1s. Plot the x - t graph of his motion. Determine graphically and otherwise how longthe drunkard takes to fall in a pit 13 m away from the start.



Sol.

It is clear from the x – t graph that the time taken by the drunkard to fall in the pit13 m away from the star = 37s.

Otherwise, it is clear that

2m 2m 2m 2m

x 5m 3m 5m 3m 5m 3 m 5m 3m 5m 13m

(taking x as positive for forward direction and negative for the backward direction)

Obviously, the number of steps involved = 4(5 + 3) + 5 = 37

and the time required for 37 step is 37 s.

Check Point 9:- A particle travels along a straight line such that the distance covered by it in anysecond, exceeds, to that in the previous second by 2m. If it covers a distance of 1m in the firstsecond. Show that its distance at the end of n sec. is n2 metres.

O 1 sec 2 sec 3 secst nd rd

D – 1 1+2 1+2+2

Sol. Evidently distance covered in nth sec. = 1+(n –1) 2 = (2n –1)

Total distance covered at the end of n sec = (2n 1) 2 n 1

n(n 1)2 n

2

= n(n +1) – n

= n2.

Check Point 10:- A point P consider at contact point of a wheel on ground which

rools on ground without sliping then value of displacement of point P when

wheel completes half of rotation - [If radius of wheel is 1 m]

Sol. Displacement = 4 2 Ans.

Speed and Velocity

Speed The rate of change of distance with time is called speed.

Total distance

SpeedTime taken



SI unit ms–1

Dimensional Formula: 0 1 -1[M L T ]

Speed is a Scalar Quantity.

Speed of an object can be positive or zero but not negative. It is because distance cannot benegative.

Average Speed (in an Interval) : It is the ratio of the total distance travelled by the object to the totaltime taken.

total distanceAverage speed

total time taken

Note : If any car covers distance S1, S

2, ........ in the time intervals t

1, t

2, ........ then.

1 2 3 n

1 2 n

S S S ...... SV

t t ...... t

Case (i) If a particle travels distances 1 2 3S , S , S , . . . with speeds 1 2 3v , v , v , . . . respectively , in

same direction

Total distance travelled = 1 2 3S S S ........

Total time taken =31 2

1 2 3

SS S........

v v v

Average speed, 1 2 3av

31 2

1 2 3

S S S .....SS S

.....v v v

Special case : If 1 2S S S ; i.e. the body covers equal distances with v1 and v

2 speeds then,

1 2av

1 2

1 2

2 v v2 S

( v v )1 1S

v v

If body covers equal distances with three different speeds then, S1 = S

2 = S

3 = S

1 2 3

1 2 2 3 3 1

1 2 3 1 2 3

3V V V3S 3V

S S S 1 1 1 V V V V V VV V V V V V

Average speed is equal to harmonic mean of individual speeds.

Case (ii) If a particle travels with speeds 1 2 3v , v , v , . . . during time intervals 1 2 3t , t , t , . . . respec-

tively.

Total distance travelled = 1 1 2 2 3 3v t v t v t .......



Total time taken = 1 2 3t t t ......

Average speed

1 1 2 2 3 3av

1 2 3

v t v t v t .....

t t t ........(ii)

Special case : If 1 2t t t , i.e. the body move with 1v and 2v speed for equal time then,

1 2

av

(v v )t

t t

1 2av

v v

2

Average speed is equal to arithmetic mean of individual speeds.

Instantaneous Speed (At an instant) : The speed of an object at a given instant of time is called itsinstantenous speed.

Instantaneoust 0

s dsspeed Lim

t dt

Velocity The rate of change of displacement with time is called velocity.

Displacement

VelocityTime taken

SI unit: ms–1

It is a vector quantity

Dimensional formula [M0LT-1]

Note : The direction of velocity is the same as that of the displacement. The displacement can bepositive, zero or negative, so the velocity of an object may be positive, zero or negative.

Note : Slope of displacement-time graph shows velocity.

Average velocity (In a Interval) : It is the ratio of displacement to the time taken.

DisplacementAverage velocity, v

Time taken

av

xv

t

Instantaneous Velocity (At an instant) : The velocity of an object at any given instant

of time is called its instantaneous velocity.

t 0

x dxv lim

t dt

Check Point 11:- In the Check Point 1, if Ram takes 4 seconds and Shyam takes 5 seconds togo from P to Q, find(a) Average speed of Ram and Shyam?(b) Average velocity of Ram and Shyam?



Sol: (a ) Average speed of Ram =4

100 m/s = 25 m/s

Average speed of Shyam = 5

50 m/s = 10 m/s

(b) Average velocity of Ram =4

100 m/s = 25 m/s (From P to Q)

Average velocity of Shyam =5

100 m/s = 20 m/s (From P to Q)

Check Point 12:- A person went to market with v1 speed and comeback with v2 speedalong a straight line. Find out the average speed and average velocity of theperson?

Sol: Let total distance travelled by the person be 2s.

Time taken to travel first part =1v

s

Time taken to travel next part =2vs

Average speed =takentimeTotalcovereddistanceTotal

=

21 vs

vs

s2

=

21

21

vvvv2

(harmonic progression)

Average velocity = 0

Check Point 13:- The displacement x in metres of a particle of mass m kg moving in one dimensions

under the action of a force, is related to the time t in seconds, given by the equation t x 3 . Find the

displacement of the particle when is velocity is zero.

Sol. As t x 3, x t 3

or x = (t – 3)2

Thus, 2dx dv t 3

dt dt

= 2 (t – 3) = 2t – 6

when v = 0, 2t – 6 = 0 or t = 3s and then,

x = (t – 3)2 = (3 – 3)2 = 0

Check Point 14:- A particle is moving along the X-axis. It's position as a function of time is given by :

2x A t B

2A 2.1 ms ; B 2.8 m

Determine (i) the displacement of the particle during the time interval from 1t = 3 s to 2t = 5

s (ii) the average velocity during this time interval and (iii) the magnitude of instantaneous velocityat t = 5 s.

Sol.(i) 2x A t B



At 1t = 3 s, the position of the particle is2

1x (2.1) (3) 2.8 21.7 m

At 2t = 5 s, the position of the particle is 22x (2.1) (5) 2.8 55.3 m

Displacement, 2 1x x x 55.3 21.7 33.6 m

(ii) Magnitude of average velocity is

v =x 55.3 21.7 33.6

16.8 m / sect 5 3 2

(iii) Instantaneous velocity :

2dx dv A t B 2At

dt dt

v = 2 × 2.1 × 5 = 21 1ms

Special Points* The magnitude of instantaneous velocity and instantaneous speed are equal.* The determination of instantaneous velocity by using the definition usually involves

calculation of derivative. We can find v = dtdx

by using the standard results from

differential calculus.* Instantaneous velocity is always tangential to the path.* Average speed is always positive in contrast to average velocity which being a

vector, can be positive or negative.* If the motion of a particle is along a straight line and in same direction then,

average velocity = average speed.* Average speed is, in general, greater than the magnitude of average velocity.

AccelerationThe rate of change of velocity is called acceleration

Acceleration =Change in velocity

Time interval

Type of change in velocity

Only direction is change Only magntiude is change Direction and magnitude both are change

acceleration is perpendicular to acceleration is parallel or acceleration has two components, one is

velocity anti parallel to velocity parallel or antiparallel and other isperpendicular

e.g. Uniform circular motion e.g. Motion under gravity e.g. Parabolic Motion



SI unit : ms–2

Acceleration is a Vector Quantity.

Dimensional formula [M0L T–2].

Direction of acceleration is same as change in velocity.

Acceleration is positive if velocity is increases and is negative if the velocity decreases. The negativeacceleration is also called retardation or deceleration.

Average acceleration It is the ratio of the total change in velocity to the total time taken.

Average acceleration

f iav

f i

v vtotal change in velocity va

total time taken t t t

Check Point 15 :- A car is moving in the positive x direction in 20 ms–1 and comes up behind a truckand is unable to pass. The car slows to 15 ms–1 in a time 2 s. Calculate car's average acceleration.

Sol.: Change in car's velocity, 22 1v v v 15 20 5 ms

Note that the change in velocity is negative because the car's velocity in the positive x direction isdecreased by 5 ms–1.

Time interval, 2 1t t t 2 s

Average acceleration, 2v 5a 2.5 ms

t 2

The average acceleration is negative because the car's velocity along the positive x direction isdecreased.

Instantaneous acceleration The acceleration of the object at a given instant of time is called itsinstantaneous acceleration.

t

v dva lim

t dt

Note: If an object is moving with uniform acceleration, then, instantaneous acceleration = uniformacceleration = average acceleration.

Check Point 16:- The position of a particle moving along a straight line is given by:

x = 2 – 5t + 6t2

Find the acceleration of the particle at t = 2s.

Sol. x = 2 – 5t + 6t2 or 2dx d2 5t 6t 5 12t

dt dt

2

2

d x da 5 12t 12

dtdt

Note: If velocity change with displacement then acceleration.

dva

dt



dv dxa .

dt dx

dva v

dx

Special Points

* If a particle is moving with uniform acceleration, this does not necessarily imply that particleis moving in straight line.

Example : Parabolic motion

* However if acceleration is constant, acceleration is uniform but motion is non-uniform and ifacceleration is not constant, both motion and acceleration are non-uniform.

* (i)d | v |

0dt

while

dv0

dt

(it is possible.) (think)

(ii)d | v |

0dt

while

dv0

dt

(it is not possible.) (think)

Properties of Graph1. Main quantity: on x-axis

2. Dependent quantity: on y-axis

3. Slope of graph:y axis

x axis

4. Area covered with x- axis by graph: y axis x axis

tan

s

vt

* Slope of distance- time graph represent speed of the object.

distance travel from time t1 to t

2 = area of AB t

2t1

area = length × width = V × (t2 – t

1)

A BV

t1 t2t* In speed-time graph, area covered with x-axis shows distance.

Average Velocity

Vav = t

x

=if

if

tt

xx

the average velocity is equal to the slope of the line (chord)

x

Q

P

tO

xf

xi

ti tf

joining the points corresponding to P and Q on the x-t(position-time) graph.



Instantaneous Velocity : The instantaneous velocity at P is the slope of the tangent atP in the x t (position-time) graph .

t 0

x dxv lim

t dt

When the slope of the x t graph is positive, v is positive (as atthe point A in figure). At C, v is negative because the tangent hasnegative slope. The instantaneous velocity at point B (turning point)is zero as the slope is zero.

Average acceleration

The slope of straight line joining two points on velcoity-time graph is the average acceleration of theobject between these two points.

The average acceleraton can be positive or negative depending

upon the sign of slope of velocity-time graph It is zero if the change invelocity of the object in the given interval of time is zero.

av

total change in velocity va

total time taken t

* Instantaneous acceleration is also the slope of the tangent to the velocity-time graph at giventime.

t

v dva lim

t dt

v

ta<0

a=0

0

a>0a=0

a=0

Some extra points about graphsStraight line-equation, graph slope (+ve, –ve, zero slope)

y = mx + c (equation of a straight line)slope = mintercept = c on the yaxis.

m = slope = tan =dy

dx

Cx

+ve slope

y

Cx

slope = 0

y

C

x

–ve slope

y



Parabolic curve-equation, graph

x

y

y = kx2 x

y

y = –kx2

x

y

x = ky2 x

y

x = –ky2

Where k is a positive constant.

Motion with uniform velocityxt graph is a straight line of slope v

x

O

xi

t v is positive

slope

=v

x

O

xi

tv is negative

slope =v

as velocity is constant, v t graph is a horizontal line.

v

O

u

t

positive velocity

v

O

u

t

negative velocity

at graph coincides with time axis because a = 0 at all time instants.

Graphs in uniformly accelerated motion (a 0)* x is a quadratic polynomial in terms of t. Hence x t graph is a parabola.

xi

x

a > 0

t0

xi

x

a < 0

t0

x-t graph



* v t graph is a straight line of slope a.

v

ua is positive

slope

= a

t0

v

u

a is negative

slope = a

t0

v-t graph* at graph is a horizontal line because a is constant.

a

apositiveacceleration

t0a

a

negativeacceleration

0

a-t graphVariable Velocity : The area between the velocity-time graph and the time-axis gives the displacement.

The v-t graph represents variable acceleration.

Variable acceleration : For a body moving with variable acceleration, the a -t graph is acurve. The area between the a -t graph and the time-axis gives the change in velocity,Change in velocity = Area 1 - Area 2 + Area 3

Special Points* For uniformly accelerated motion (a 0), xt graph is a parabola (opening

upwards if a > 0 and opening downwards if a < 0). The slope of tangent atany point of the parabola gives the velocity at that instant.

* For uniformly accelerated motion (a 0), vt graph is a straight line whose slopegives the acceleration of the particle.

* In general, the slope of tangent in xt graph is velocity and the slope of tangentin vt graph is the acceleration.

* The area under at graph gives the change in velocity.* The area between the vt graph gives the distance travelled by the particle, if

we take all areas as positive.* Area under vt graph gives displacement, if areas below the taxis are taken

negative.



Some more Graphs(i) Non uniform velocity (ii) Non uniform velocity (decreasing with

time)

(increasing with time) In this case; In this case;As time is increasing, is also As time increases, decreases.increasing.

dtdx

= tan is also increasing dtdx

= tan also decreases.

Hence, velocity of particle is Hence, velocity of particle is decreasing.increasing.

(iii) Uniform acceleration (iv) Uniform retardation

tan is constant. Since > 90º

dtdv

= constant tan is constant and negative.

Hence, it shows constant acceleration. dtdv

= negative constant

Hence, it shows constant retardation.(v) Uniformly increasing acceleration (vi) Uniformly decreasing acceleration

is constant. Since > 90º0º < < 90º tan > 0 tan is constant and negative.

dtda

= tan = constant > 0 dtda

= negative constant

Hence, acceleration is uniformly Hence, acceleration is uniformlyincreasing with time. decreasing with time



(vii) For a body projected upwards, When the (viii) For a ball dropped on the ground frombody moves, its speed decreases uniformly, a certain height, As the ball falls, its speedbecoming zero at the highest point. As increases. As the bn all bounces back, itsthe body moves down, its speed increases speed decreases uniformly and becomes zerouniformly. It returns with the same speed at the highest point.with which it was thrown up.

(ix) The v-t graph represents a body projected upwards with an initial velocity u. The velocitydecreases with time (negative uniform acceleration), becoming zero after certain time t. Thenthe velocity becomes negative and increases in magnitude, showing body is returning tooriginal position with positive uniform acceleration.

Note:2vdv 1 dv

a = =ds 2 ds

This may also be written asdv a

= .ds v

That means, derivative of 'v' with

displacement 's' yields the ratio of acceleration and speed.

Velocity - displacement Graph

Acceleration = speed × slope of v–s graph =1

2× slope of v2–s graph.

The slope can be positive, negative or zero which signify positive, negative or zero accelerations.

Acceleration-displacement graph

2

0 1

v s

v svdv = a.ds

2

1

s2 20 s

v = v +2 a.ds

20v = v +2(areaunder a - s graph)



Some Imposible Graphs

Uniformly Accelerated Motion: If a particle is accelerated with constant acceleration

in an interval of time, then the motion is termed as uniformly accelerated motionin that interval of time.For uniformly accelerated motion along a straight line (xaxis) during a timeinterval of t seconds, the following important results can be used.

(a) tuva

(b)2

uvVav

(c) S = (vav)t

(d) t2

uvS

(e ) v = u + at

(f ) s = ut + 1/2 at2

s = vt 1/2 at2

xf = x

i + ut + 1/2 at2



(g) v2 = u2 + 2as

(h) sn = u + a/2 (2n 1)

u = initial velocity (at the beginning of interval)a = accelerationv = final velocity (at the end of interval)s = displacement (x

f x

i)

xf = final coordinate (position)

xi = initial coordinate (position)

sn = displacement during the n th sec

Directions of Vectors in Straight Line MotionIn straight line motion, all the vectors (position, displacement, velocity & acceleration)will have only one component (along the line of motion) and there will be onlytwo possible directions for each vector.

* For example, if a particle is moving in a horizontal line (xaxis), the twodirections are right and left. Any vector directed towards right can be representedby a positive number and towards left can be represented by a negative number.

line of motion

line

ofm

otio

n

line

ofm

otio

n

+

+ +

-

- -

* For vertical or inclined motion, upward direction can be taken +ve and downwardas ve

* For objects moving vertically near the surface of the earth,the only force actingon the particle is its weight (mg) i.e. the gravitational pull of the earth. Henceacceleration for this type of motion will always be a = g i.e. a = 9.8 m/s2 (ve sign, because the force and acceleration are directed downwards, If weselect upward direction as positive).

Note : If acceleration is in same direction as velocity, then speed of the particle

increases.* If acceleration is in opposite direction to the velocity then speed decreases i.e.

the particle slows down. This situation is known as retardation.

Special Points

* Actually first and second equations of motion are vector equations while third one is a scalarequation, i.e.,

v u at

21

2s u t a t

2 2, . 2 , 2 .v v u u a s or v u a s



* If the velocity and acceleration are collinear.

2

2 2

1

22

u at

s ut at

u as

* If acceleration and velocity are not collinear.2 2 1/2[ ( ) 2 cos ]u at uat

Check Point 17:- A particle starts moving from the position of rest under a constant acceleration. If ittravels a distance x in t sec, what distance will it travel in next t sec?

Sol:As ace. is constt., from 2nd equation of motion,

21

2x at [as u = 0] ...(1)

Now if it travels a distance y in next t sec., the total distance travelled in (t + t = 2t) secwill be x + y, so

21(2 )

2x y a t ...(2)

eq. 2 - eq. 12 21 1

42 2

x y x a t at

213

2y at

3y x

Check Point 18:- If a car is move with 0( )v initial velocity. Calculate distance travelled by car after the

brakes applied.

Solu.

2 2 2v u as

When car come to rest 0v

0u v

200 2v as

20

2

vs

a

Check Point 19:- The velocity acquired by a body moving with uniform acceleration is 20 m/s in first2 sec and 40 m/s in first 4 sec. Calculate initial velocity.

Sol.2 1

2 1

V Va

t t

240 20 20a 10m / s

4 2 2

Now, v = u + at



v1 = u + at

1

20 = u + 10 × 2

20 = u + 20 u = 0 m/s

Check Point 20:- A particle starts with initial velocity 2.5 m/s along the x direction and acceleratesuniformly at the rate 50 cm/s2. Find time taken to increase the velocity to 7.5 m/s.

Sol. v = u + at

7.5 = 2.5 + 0.5 t

5.0 = 0.5t

50t = =10sec

5

Check Point 21:- A particle starts with a constant acceleration. At a time t second speed is found to be100 m/s and one second later speed becomes 150 m/s. Find acceleration of the particle.

Sol. From equation (1) of motion v = u + at

100 = 0 + at

100 = at ...(1)

Now consider velocity one second later-

v' = 0+a(t +1)

150 = a(t+1) ...(2)

On subtracting equation (1) from equation (2)

a = 5 0 m / s

2

Check Point 22:- G

Sol. (a)2

T

1S = at

2

2T

1S = (1.5)t

2...(1)

Dostance covered by car when car overtakes the truck

2c

1S = (2)t

2

2T

1(S +150) = (2)t

2...(2)

Divide equation (2) by equation (1)T

T

S +150 2=

S 1.5 T

150 20 41+ = =

S 15 3

T

150 4 1= –1=

S 3 3 or ST = 450

Distance travelled by car = 450 + 150 = 600 metre



(b) Now by equation (1)2

T

1S = at

2

21450= 1.5 t

2

2 450 2t =

1.5 t = 300 2 =24.5sec

Therefore car will overtake the truck after 24.5 second.

Check Point 23:- A body travels a distance of 2 m in 2 seconds and 2.2 m in next 4 seconds. Whatwill be the velocity of the body at the end of 7th second from the start.

Sol. Here, case (i) s = 2m, t = 2s

case (ii) t = 2 + 4 = 6s

Let u and a be the initial velocity and uniform acceleration of the body.

we know that,21

s = ut+ at2

case (i) × × 212= u 2+ a 2

2

or 1 = u + a ...(1)

case (ii) × × 214.2= u 6+ a 6

2

or 0.7 = u + 3a ...(2)

subtracting (2) from (1), we get

0.3 = 0 – 2a = –2a

or a = – 0.3/2 = – 0.15 ms–2

From (i), u = 1 – a = 1 + 0.15

u = 1.15 ms–1

For the velocity of body at the end of 7th second, we have

u = 1.15 ms–1; a = – 0.15 ms–2

v = ?, t = 7s

As, v = u + at

v = 1.15 + (–0.15) × 7

v = 0.1 m/s

Check Point 24:- A body travels a distance of 20m in the 7th second and 24 m in 9th second. Howmuch distance shall it travel in the 15th second?

Sol. Here, S7= 20 m ; S

9 = 24 m,

S15 = ?

Let u and a be the initial velocity and uniform acceleration of the body.

n

aS = u+ (2n -1)

2



×7

aS = u+ (2 7 –1)

2

13a20= u+

2...(i)

and ×g

aS = u+ (2 9 -1)

2

1724= u+ a

2...(ii)

Subtracting (ii) form (i), we get

4 = 2a

a = 2 ms–2

Putting this value in (i), we get

×13

20= u+ 22

or 20 = u + 13

u = 20 – 13 = 7 ms–1

Hence, × ×15

a 2S = u+ (2 15 –1) =7+ 29

2 2

S15= 36 m Ans.

Check Point 25:- A driver takes 0.20 s to apply the brakes after he sees a need for it. This is calledthe reaction time of the driver. If he is driving a car at a speed of 54 km/h and the brakes causea deceleration of 6.0 m/s2, find the distance travelled by the car after he sees the need to putthe brakes on.

Sol. Distance covered by the car during the application of brakes by driver-

u = 54 km/h = ×5

54 m/s =15m/s18

S1 = ut or S

1= 15 × 0.2 = 3.0 meter

After applying the brakes;

v = 0 u = 15 m/s, a = 6 m/s2 S2= ?

v2 = u2 – 2as

0 = (15)2 – 2 × 6 × S2

12 S2= –225

2

225S = =18.75metre

12

Distance travelled by the car after driver sees the need for it

S = S1+ S

2

S = 3 + 18.75 = 21.75 metre. Ans.

Check Point 26:- A particle moving rectilinearly with constant acceleration is havinginitial velocity of 10 m/s. After some time, its velocity becomes 30 m/s. Find



out velocity of the particle at the mid point of its path?Sol : Let the total distance be 2x.

distance upto midpoint = xLet the velocity at the mid point be vand acceleration be a.From equations of motion

v

2 = 102 + 2ax ____ (1)2 = v2 + 2ax ____ (2)

(2) - (1) givesv2 - 302 = 102 - v2

v2 = 500 v = 510 m/s

Check Point 27:- A police inspector in a jeep is chasing a pickpocket an a straightroad. The jeep is going at its maximum speed v (assumed uniform). Thepickpocket rides on the motorcycle of a waiting friend when the jeep is at adistance d away, and the motorcycle starts with a constant acceleration a. Show

that the pick pocket will be caught if 2adv .

Sol : Suppose the pickpocket is caught at a time t after motorcycle starts. Thedistance travelled by the motorcycle during this interval is

2at21s ____ (1)

During this interval the jeep travels a distance

vtds ____ (2)

By (1) and (2),

vtdat2

1 2

or,a

ad2vvt2

The pickpocket will be caught if t is real and positive.This will be possible if

2adv2 or, 2adv Check Point 28:- A man is standing 40 m behind the bus. Bus starts with 1 m/sec2

constant acceleration and also at the same instant the man starts moving withconstant speed 9 m/s. Find the time taken by man to catch the bus.

x = 0t = 0

40 m t = 0x = 40

1m/sec2

Sol: Let after time ‘t’ man will catch the busFor bus

x = x0 + ut +

21 at2 , x = 40 + 0(t) +

21 (1) t2



x = 40 +2t2

............. (i)

For man, x = 9t ............. (ii)From (i) & (ii)

40 +2t2 = 9t or t = 8 s or t = 10s.

Check Point 29 :- A truck and a car are brought to a hault by application of same braking force.Which one will come to stop in a shorter distance if they are moving with same (a) velocity(b) kinetic energy and (c) momentum?

Sol : By applying brakes the body is brought to rest, so 0 and a = (-F/m) (as it is

retardation). Ifs is the distance travelled in stopping (called stopping distance), from 3rd equationof motion,

2 2 2u as 20 2( / )u F m s

2

2

mus

F

21

2KE mu and also

2

2

pKE

m

2 2

2 2

mu KE ps

F F mF

(a) If u is same:2( / 2 )s mu F , i.e., s m

as mass of car is less than that of truck, so car will stop in shorter distance.(b) If KE is same:

( / )s KE Fso both will stop after travelling same distance.

(c) If p is same:

s = (p2/2mF), i.e., (1 / )s mAs mass of truck is more than that of car, so truck will stop in a shorter distance.

Check Point 30:- A car accelerates from rest at a constant rate for sometime after which it

decelerates at constant rate to come to rest. If the total time elapsed is t second, evaluate (i)

the maximum velocity reached and (ii) the total distance travelled.

Sol. Le t1 be the time during which the car accelerates and t

2 be the time for its deceleration.

(i) Let vm be the maximum velocity reached,

v = v0 + at,

for timem

1 m 1

vt , v 0 t, or t

for timem

2 m 2 2

vt , 0 v t or t

As t1 + t

2 = t



m mv vt

or mv t

m

tv

(ii)2

0

1s v t at

2

for time2

1 1 11

t , s t2 0 1v v, a , t t

and for time 22 2 m 2 2

1t , s v t t

2

Total distance covered, s = s1 + s

2

22

m m mm

v v v1 1s v

2 2

2 2 2 2 22m m m m mm

v v v v v 1 1 1s v

2 2 2 2 2

2 2 2 2

2

1 t ts

2 2

Alternative Solution : t = t1 + t

2

slope of OA curve = tan = =1

max

tv

slope of AB curve = =2

max

tv

vmax

O

t2t1

B

AV

tt = t1 + t

2

t =maxv

+ maxv

vmax = t

Total distance travelled by the car in time ‘t’

Area under graph (directly) = )(t

21 2

= )(2t2

Ans.

Motion of an Object Under Free FallIf an object is released from a height near the surface of earth, it is accelerated downwards underthe influence of a gravity pull, with acceleration due to gravity, g. If the air resistance isneglected.The value of g near the surface of earth is generally taken equal to 9.8 ms-2. It means,free falls is a case of motion with uniform acceleration.



1. If a body is free falling from height (Object is released from the rest)

(i) Time required to reach body at the earth

u = 0

v

h

g

v

g

ht

2

ghv 2

g

vh

2

2

Fig. 2.12

21

2h gt

2ht

g

(ii) Final velocity at the earth

2 2 2V u gh ¼ 0u ½

2V gh

2

2

vh

g

u = 0 (Initially body is in rest ½

a = + g (direction of acceleration is same as in motion½

v = gt ...(i)

21

2h gt ...... ii

2 2v gh ...... iii

Distance travelled in nth second

2 12n

gh n ..... iv

Graphs of displacement, velocity and acceleration with time t

If downward direction taken negative, so g becomes negative then graphs of displacement, velocityand acceleration with time.



Final Velocity v = gt v tAt t = 0 v = 0, t = 1 sec v = 10 m/sec, t = 2 sec

v = 20 m/sec etc.

* 21

2h gt or 2 ,h t so distances travel in time , 2 ,3 ,t t t etc.

are 2 2 31 : 2 : 3 , respectively.

* Distance travel in nth second 12 1

2nh g n

so distance travel in first, second and third seconds arerespectively 1 : 3 : 5.

Galileo law of odd numbers. The distances traversed,during equal intervals of time, by a body falling from rest,stand to one another in the same ratio as the odd

numbers beginning with unity (1:3:5:7......)

2. If body a is free falling (Object is released with some initial

velocity)

By equation of motion

v u gt

21

2h ut gt

2 2 2v u gh

2 12n

gh u n

3. If Body is through Vertically Upwards

* Body is move upward and acceleration is downward, so a = – g

(A) velocity of body after time t

V = u – gt

(B) height of body after time t

21

2h ut gt

(C) Velocity at h height

2 2 2V u gh

(D) Distance travelled in nth second

12 1

2nh u g n

* for maximum height 0v

0 u gt



u gt

ut

g

21h ut gt

2

2 21h gt gt

2

21h gt

2

2ht

g

2 20 u 2gh

u2 = 2gh

u 2gh

* graphs of displacement and velocity with time

* Time required to move upward is equal to the time required for

move downward under the gravity.

Total flying time 1 22

u u u

T t tg g g

Check Point 30:- From the top of a tower a stone is thrown up which reaches the ground in time t1. Asecond stone thrown down with the same speed reaches the ground in a time t2. A third stonereleased from rest from the same location reaches the ground in a time t3. Then:

(A)3 1 2

1 1 1= +

t t t (B) 2 2 23 1 2t = t – t (C) 1 2

3

t + tt =

2(D) 3 1 2t = t t

Sol. For first case2

1 1

1-h= ut - gt

2



t1

u u u = 0

t2 t3

h h h

21 1

1h= ut + gt

2...(1)

For second case2

2 2

1–h= –ut – gt

2

22 2

1h= ut + gt

2...(2)

For third case2

3

1–h= – gt

2

23

1h= gt

2...(3)

equation (1) × t2+ equation (2) × t

1

1 2 1 2 1 2

1h(t + t ) = gt t (t + t )

2

From equation (3)23 1 2

1 1gt = gt t

2 2

3 1 2t = t t

Check Point 31:- A ball is dropped from the top of a building of height h. At the same time anotherball is thrown vertically upward with velocity u from the ground. At which height and what timetwo balls meet.

Sol. For first particle21

–x = – gt2

21x = gt

2...(1)

u = 0

u

x

h – x

For second particle21

h - x = ut – gt2

Value of x from equation (1)

2 21 1h – gt ut – gt

2 2



h = ut

ht =

u

From equation (1)2

2

1 hx = g

2 u

From the surface = h – x

Check Point 32:- A juggler throws balls into air. He throws one whenever the previous one is at itshighest point. How high do the balls rise if he throws n balls each sec? Acceleration due to gravityis g.

Sol. As the juggler is throwing n balls each second and 2nd when the first is at its highest point,so the time taken by one ball to reach the highest point,

t = (1/n) s

and as at highest point 0, from 1st equation of motion,

0 = u – (g) (1/n), i.e., u = (g/n) ...(1)

Now from 3rd equation of motion, i.e., 2u 2as, 2v

0 = u2 – 2gh, i.e., h = (u2/2g)

or 2

g gh= as fromEqn.(1),u =

n2n Ans.

Check Point 33:- A body falling freely from a given height H hits an inclined plane in its path at aheight 'h', As a result of this impact the direction of the velocity of the body becomes horizontal.For what value of (h/H) the body will take maximum time to reach the ground?

Sol. In accordance with fig. time taken by the body to strike the inclined plane

1

2(H – h)t =

g

Now, as after impact the velocity of the body is horizontal,

so time taken to reach the ground

2

2ht =

g

So total time of motion 1 2

2t = t + t = [ h + (H – h) ]

g

For t to be maximum (dt/dh) = 0

i.e., 0

1/2 1/2d 2{h + (H - h) }

dh g



or–1/2 –1/21 1

h + (H - h) (–1) =02 2

0

2as

g

or h = H – h, i.e.,h 1=

H 2 Ans.

Check Point 34:- A particle is dropped from a tower. It is found that it travels 45 min the last second of its journey. Find out the height of the tower? (takeg = 10 m/s2)

Sol: Let the total time of journey be n seconds.

Using; )12(2

na

usn 45 = 0 +2

10)12( n

n = 5 sec

Height of tower; h =2

1gt2 =

2

1 10 52 = 125 m

Check Point 35:- A ball is projected vertically up with an initial speed of 20 m/s on a planet whereacceleration due to gravity is 10 m/s2

(a) How along does it takes to reach the highest point?

(b) How high does it rise above the point of projection?

(c) How long will it take for the ball to reach a point 10m above the point of projection?

Sol. As here motion is vertically upwads,

a = –g and v = 0

(a) From 1st equation of motion, i.e, v = u + at,

0 = 20 – 10t

i.e. t = 2 sec.

(b) Using v2 = u2 + 2as

0 = (20)2 – 2 × 10 × h

i.e. h = 20 m.

(c) Using s = ut +1

2at2

× × 2110=20t (– ) 10 t

2

i.e. t2 – 4t + 2 = 0

or t =2± 2,

i.e. t = 0.59 sec. or 3.41 sec.

i.e., there are two times, at which the ball passes through h = 10 m, once while going upand then coming down.

Check Point 36:- A rocket is fired vertically up from the ground with a resultant vertical accelration of10 m/s2. The fuel is finished in 1 minute and it continues to move up.

(a) What is the maximum height reached?



(b) After how much time from then will the maximum height be reached? (Take g = 10 m/s2)

Sol. (a) The distance travelled by the rocket during burning interval (1 minute = 60 s) in whichr e s u l t a n t a c c e l e r a t i o n i s v e r t i c a l l y u p w a r d s a n d 1 0 m / s

2 will be

h1 = 0 × 60 + (1/2) × 10 × 602 = 18000 m ...(1)

Velocity acquired by it is

v = 0 + 10 × 60 = 600 m/s ...(2)

After one minute the rocket moves vertically up with initial velocity of 600 m/s andcontinues till height h

2till its velocity becomes zero.

0 = (600)2 – 2gh2

or h2= 18000 m [as g – 10m/s2] ...(3)

From equations (1) and (3) the maximum height reached by the rocket from the ground is

H = h1+ h

2= 18 + 18 = 36 km

(b) The time to reach maximum height after burning of fuel is

0 = 600 – gt

t = 60 s

After finishing fuel the rocket goes up for 60 s.

Check Point 37:- A body is released from a height and falls freely towards the earth. Exactly 1 seclater another body is released. What is the distance between the two bodies after 2 sec therelease of the second body, if g = 9.8 m/s2.

Sol. The 2nd body falls for 2s, so

22

1h = g(2)

2...(1)

While 1st has fallen for 2 + 1 = 3 sec so

21

1h = g(3)

2...(2)

Separation between two bodies after 2 sec the release of 2nd body,

2 21 2

1d =h – h = g(3 - 2 )

2 = 4.9 × 5 = 24.5 m

Check Point 38:- A stone is dropped from a balloon going up with a uniform velocity of 5m/s. If the balloon was 60 m high when the stone was dropped, find itsheight when the stone hits the ground. Take g = 10 m/s2.

Sol: S = ut +21 at2

– 60 = 5(t) +21 (–10) t2

– 60 = 5t – 5t2

5t2 – 5t – 60 = 0

60m

+ve

–vet2– t – 12 = 0t2 – 4t + 3t – 12 = 0(t – 4) (t + 3) = 0

t = 4



Height of balloon from ground at this instant= 60 + 4 5 = 80 m

Check Point 39:- Two stones are thrown up simultaneously from the edge of a cliff 200 m high with

initial speeds of 15 m/s and 30 m/s. Verify that the following graph correctly represents the

time variation of the relative position of the second stone with respect to the first. Neglect air

resistance and assume that the stone do not rebound after hitting the ground. Take g = 10 m/s2.

Give the equations for the linear and curved parts of the plot.

Sol.2

0 01

x x v t gt2

For the first stone, 20 0 0x 200mv , v 15m / s, g 10m / s

21x 200 15t 5t ...(i)

From the second stone, 0x 200m

20v 30 m / s, g 10m / s

22x 200 30t 5t ...(iii)

When the first stone hits the ground, x1 = 0 and as such that

20 200 15t 5t

2t 3t 40 0

t 8 t 5 0

either t = 8 s or t = -5 s

Since t = 0 corresponds to the instant when the stone was projected, negative time (t = 5s)

has no meaning in this case.

Check Point 40:- A stone is dropped into a well in which the level of water is h below the top of thewell. If v is velocity of sound, the time T after which the splash is heard is given by

Sol. Suppose, t1 = time taken by stone to reach the level of water

t2 = time taken by sound to reach the top of well

so, T = t1 + t

2

For t1 : u = 0

h = ut + 21 gt2 h = 0 + 2

1 gt

12 t

1 = g

h2

For t2 : As the velocity of sound is constant



h = Vt2 t

2 = V

h

Therefore, T = Vh

gh2

Motion with non-uniform acceleration

x = f

i

t

t

dt)t(v (displacement in time interval t = ti to t

f)

v = v f v i = f

i

t

t

dt)t(a

Check Point 41:- An object starts from rest at t = 0 and accelerates at a rate given by a =6t. What is

(a) its velocity and(b) its displacement at any time t?

Sol: t

t

v(t)

)v(t 00

a(t)dtdv

Here t0 = 0 and v(t

0) = 0

v(t) = t

0

6tdt = 0

t

2

t6

2

= 6 (

2

t2

- 0) = 3t2

So, v(t) = 3t2

As t

t0

v(t)dtx

t

0

2dt3tx = 0

t

3

t3

3

=

0

3

t3

3

= t3

Hence, velocity v(t) = 3t2 and displacement 3tx .

Check Point 42:- For a particle moving along x-axis, acceleration is given as 22va . If thespeed of the particle is v0 at x = 0, find speed as a function of x.

Sol: a = 2v2 or dtdv

= 2v2 or dxdv

× dtdx

= 2v2

v dxdv

= 2v2 dxdv

= 2v

v

v0v

dv =

x

0

dx2 vv0nv = x0x2

0vvn = 2x v = v

0e2x Ans.

Check Point 43:- The acceleration experienced by a moving boat, after its engine is cut-off, is given bydv/dt = – kv3 where k is a constant. If v0 is the magnitude of the velocity at cut-off, find themagnitude of the velocity at a time t after the cut-off.

Sol. 3dvkv

dt



3

dvkdt

v

Integrating both sides within proper limits,

0

v t

3v 0

dvk dt

v

0

vt02

v

1k | t |

2v

2 20

1 1kt

2v 2v

20

2 2 20 0

1 2v kt1 1kt

2v 2v 2v

22 0

20

vv

1 2v kt

0

20

vv

1 2v kt

Check Point 44:- For a particle moving along + x-axis, acceleration is given as a = x. Findthe position as a function of time? Given that at t = 0 , x = 1 v = 1.

Sol: a = x dxvdv

= x 2v2

=2x2

+ C

t = 0, x = 1 and v = 1 C = 0 v2 = x2

v = ± x but given that x = 1 when v = 1

v = x dtdx

= x x

dx= dt

nx = t + C 0 = 0 + C nx = tx = et

Check Point 45:- For a particle moving along x-axis, acceleration is given as a = v. Findthe position as a function of time ?Given that at t = 0 , x = 0 v = 1.

Sol: a = v dtdv

= v vdv

= dt

nv = t + c 0 = 0 + c

v = et dtdx

= et dx = dtet

x = et + c 0 = 1 + cx = et – 1



Critical Thinking Questions

Check Point 46:- A particle covers43

of total distance with speed v1 and next41

with v2.

Find the average speed of the particle?

Sol: Let the total distance be s3s/4

AB

Cs/4

average speed (< v >) = takentimeTotalcetandisTotal

< v > =

21 v4s

v4s3

s

=

21 v41

v43

1

=

21

21

v3vvv4

Check Point 47:- A car is moving with speed 60 Km/h and a bird is moving with speed90 km/h along the same direction as shown in figure. Find the distance travelled bythe bird till the time car reaches the. tree?

Sol: Time taken by a car to reaches the tree

240 m(t) = hr/km60

m240 = hr

6024.0

Now, the distance travelled by the bird during this time interval (s)

= 6024.090 = 0.12 × 3 km = 360 m.

Check Point 48:- The position of a particle moving on X-axis is given byx = At3 + Bt2 + Ct + D.

The numerical values of A, B, C, D are 1, 4, -2 and 5 respectively and SI units areused. Find (a) the dimensions of A, B, C and D, (b) the velocity of the particle at t = 4s, (c) the acceleration of the particle at t = 4s, (d) the average velocity during theinterval t =0 to t = 4s, (e) the average acceleration during the interval t = 0 to t = 4 s.

Sol: As x = At3 + Dt2 + Ct + D

(a) Dimensions of A, B, C and D ,[At3] = [x] (by principle of homogeneity)

[A] = [LT–3]

similarly, [B] = [LT -2], [C] = [LT -1] and [D] = [L] ;

(b) As v = dtdx

= 3At2 + 2Bt + C

velocity at t = 4 sec.

v = 3(1) (4)2 + 2(4) (4) – 2 = 78 m/s.

(c) Acceleration (a) = dtdv

= 6At + 2B ; a = 32 m/s2

(d) average velocity as x = At3 + Bt2 + Ct = D position at t = 0, is x = D = 5m.

Position at t = 4 sec is (1)(64) + (4)(16) – (2) (4) + 5 = 125 m

Thus the displacement during 0 to 4 sec. is 125 – 5 = 120 m

< v > = 120 / 4 = 30 m/s



(e) v = 3At2 + 20 t + C , velocity at t = 0 is c = – 2 m/s

velocity at t = 4 sec is 78 m/s < a > =12

12

ttvv

= 20 m/s2

Check Point 49:- For a particle moving along x-axis, velocity is given as a function of timeas v = 2t2 + sin t. At t = 0, particle is at origin. Find the position as a function of time?

Sol: v = 2t2 + sin t dtdx

= 2t2 + sin t

x

0

dx = dt)tsint2(t

0

2 = 1tcost32x 3 Ans.

Check Point 50:- A car decelerates from a speed of 20 m/s to rest in a distance of 100m. What was its acceleration, assumed constant?

Sol: v = 0 u = 20 m/s s = 100 m as v2 = u2 + 2 as

0 = 400 + 2a × 100 a = – 2 m/s

acceleration = 2 m/s2 Ans.Check Point 51:- A 150 m long train accelerates uniformly from rest. If the front of the train

passes a railway worker 50 m away from the station at a speed of 25 m/s, what will be thespeed of the back part of the train as it passes the worker?

So: v2 = u2 + 2as

25 × 25 = 0 + 100 a

a =4

25m/s2

Now, for time taken by the back end of the train to pass the worker

we have v´2 = v2 + 2al = (25)2 + 2 ×4825

× 150

v´2 = 25 × 25 × 4

v´ = 50 m/s. Ans.Check Point 52:- A particle is thrown vertically with velocity 20 m/s . Find (a) the dis-

tance travelled by the particle in first 3 seconds, (b) displacement of the particle in 3seconds.

Sol: Highest point say B

VB = 0

v = u + gt

0 = 20 – 10 t

t = 2 sec.

distance travel in first 2 seconds.

s = s(t =0 to 2sec) + s (2sec. to 3sec.)

s = [ut + 1/2 at2]t = 0 to t = 2s

+ [ut +1/2at2]t = 2 to t = 3s

s = 20 × 2 – 1/2 × 10 × 4 + 1/2 × 10 × 12

= (40 – 20) + 5 = 25 m.and displacement = 20 – 5 = 15 m.



Check Point 53:- The displacement vs time graph of a particle movingalong a straight line is shown in the figure. Draw velocity vstime and acceleration vs time graph.

Upwards direction is taken as positive, downwardsdirection is taken as negative.

Sol: (a) The equation of motion is : x = –8t2

v = dtdx

= – 16 t ; this shows that velocity is directly pro-

portional to time and slope of velocity-time curve is negative i.e. – 16.

Hence, resulting graph is (i)

(b) Acceleration of particle is : a = dtdv

= –16.

This shows that acceleration is constant but negative.

Resulting graph is (ii)

Check Point 54:- Draw displacement–time and acceleration–time graph for the given veloc-ity–time graph.

Sol: Part AB : v-t curve shows constant slopei.e. constant acceleration or Velocity increases at constant rate with time.

Hence, s-t curve will show constant increase in slope

and a-t curve will be a straight line.

Part BC : v-t curve shows zero slope i.e. constant velocity. So, s-t curvewill show constant slope and acceleration will be zero.

Part CD : v-t curve shows negative slope i.e. velocity is decreasing with

time or acceleration is negative.

Hence , s-t curve will show decrease in slope becoming zero in the end.

and a-t curve will be a straight line with negative intercept on y-axis.

RESULTING GRAPHS ARE:



Check Point 55:- For a particle moving along x-axis, following graphs are given. Find thedistance travelled by the particle in 10 s in each case?

Sol: (a) Distance area under the v - t curve

distance = 10 × 10 = 100 m Ans.(b) Area under v – t curve

distance =21

× 10 × 10 = 50 m Ans.

Check Point 56:- The velocity of any particle is related with its displacement As; ,x = v +1

Calculate acceleration at x = 5 m.

Sol. x = v +1 2x = v +1 –12v = x

dva = v

dx

2a = x –1 2x

at x = 5 m a = 2 × 5 (25 – 1) = 240 m/s2.Check Point 57:- A toy plane P starts flying from point A along

a straight horizontal line 20 m above ground level startingwith zero initial velocity and acceleration 2 m/s2 as shown.At the same instant, a man P throws a ball vertically upwardswith initial velocity 'u'. Ball touches (coming to rest) thebase of the plane at point B of plane's journey when it isvertically above the man. 's' is the distance of point Bfrom point A. Just after the contact of ball with the plane,acceleration of plane increases to 4 m/s2. Find:( i ) Initial velocity 'u' of ball.(ii) Distance 's'.(iii) Distance between man and plane when the man

catches the ball back. (g = 10 m/s2)Sol. ( i ) Maximum height reached by ball = 20 m.

So, taking upward direction as positive, v2 = u2 + 2asso, 0 = u2 – 2 × 10 × 20 or u = 20 m/sec Ans.Also time taken by ball = t = u/g = 20/10 = 2 sec. (for touching the plane)

(ii) Horizontal distance travelled by plane in this timet = s = u

xt + 1/2 a

x t2

where, ux = initial velocity of plane,

ax = acceleration of plane.

So, s = 0 × 2 + 1/2 × 2 × 22 = 4 m(iii) Man catches the ball back 2 seconds after it

touches the plane.Velocity of plane when ball touches it v

x = u

x + a

xt = 0 + 2 × 2 = 4 m/sec.

Now, acceleration of plane becomes : ax´ = 4 m/sec2



so, s

x´ = horizontal distance travelled by plane after touch with ball = u

x´ + 1/2 a

x´t2

= 4 × 2 + 1/2 × 4 × 4= 8 + 8 = 16 m

Final distance between man and plane = s = 22 )16()20( = 656 m

Check Point 58:- The two ends of a train moving with a constant acceleration pass a certainpoint with velocities u and v. Show that the velocity with which the middle point of the

train passes the same point is2

vu 22.

Sol.

Lu

pointv

v´L2

L =a2uv 22

...(1)

2L =

a2u´v 22

...(2)

dividing (1) / (2)

2 = 22

22

u´vuv

2v´2 – 2u2 = v2 – u2

v´ =2

vu 22

Check Point 59:-A lift starts from the top of a mine shaft and descends with a constant speed of10 m/s. 4 s later a boy throws a stone vertically upwards from the top of the shaft with aspeed of 30 m/s. Find when and where stone hits the lift. [ Take: g = 10 m/s² ]

Sol.10 m/s

10 m/s

u = 30m/s

Let the time t after thrown stone it hits the lift at depth d below the top of shaft d

= ut +21 gt2

d = – 30 t +21 (10) t2 ...(1)

for liftd = 40 + 10 t ...(2)(1) = (2)– 30 t + 5 t2 = 40 + 10 t

5t2 – 40 t – 40 = 0



t2 – 8t – 8 = 0

232648t

=2

968 =

2648

t = 4 + 62Net time after to hit the lift start desending

= 4 + 4 + 62 = 8 + 62 secPutting value of t in equation (2)

d = 40 + (4 + 62 ) 10

= 40 + 40 + 620= 129 m

Check Point 60:- On a 2lane road, car A is travelling with a speed of 36 km/h. Two cars B andC approach car A in opposite directions with a speed of 54 km/h each. At a certaininstant, when the distance AB is equal to AC, both being 1 km, B decides to overtake Abefore C does. What minimum acceleration of car B is required to avoid an accident?

Sol.

B

A

C

54 km/hr54 km/hr

36 km/hr

Time taken by C to overtake A is = h/km)5436(km1

= 901 hr = 3

2× 60 sec. = 40 sec.

Let the minimum acceleration of B is a to overtake A before C then

S = ut +21 at2

1 km = (54 – 36) × 901 + 29090

1

× a

290901

× a = 1 – 9018

= 9072

a = 9072

× 90 × 90 × 2 km/h2

60606060100029070

m/sec2 = 1 m/sec2 .

Documents

Particle Dynamics 1 C O N S E R V A T IO N L A W...Target Course MASTERS ACADEMY 1-Anasagar Circular Road, Opp. Chaupati, Ajmer. 0145-2633111 Particle Dynamics 3 C O N S E R V A T