RECTILINER MOTION – JEE (MAIN+ADVANCED)

RESONANCE RECTILINEAR MOTION - 1

MECHANICS

Mechanics is the branch of physics which deals with the cause and effects of motion of a

particle, rigid objects and deformable bodies etc. Mechanics is classified under two streams

namely Statics and Dynamics. Dynamics is further divided into Kinematics and Kinetics.

1 . MOTION AND REST

Motion is a combined property of the object and the observer. There is no meaning of rest or

motion without the observer. Nothing is in absolute rest or in absolute motion.

An object is said to be in motion with respect to a observer, if its position changes with

respect to that observer. It may happen by both ways either observer moves or object moves.

2 . RECTILINEAR MOTION

Rectilinear motion is motion, along a straight line or in one dimension. It deals with the

kinematics of a particle in one dimension.

2.1 Posi t ion

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?�

2.2 Disp lacement

The change in the position of a moving object is known as

displacement. It is the vector joining the initial position ( 1r

) of

the particle to its final position ( 2r

) during an interval of time.

Displacement can be negative positive or zero.

r2

2.3 Distance

The length of the actual path travelled by a particle during a given time interval is called

as distance. The distance travelled is a scalar quantity which is quite different from

displacement. In general, the distance travelled between two points may not be equal to

the magnitude of the displacement between the same points.

RECTILINEAR MOTION


Example 1.

Ram takes path 1 (straight line) to go from P to Q and

Shyam takes path 2 (semicircle).

(a) Find the distance travelled by Ram and Shyam?P Q

1

2

100 m

(b) Find the displacement of Ram and Shyam?

Solution :

(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

2.4 Average Velocity (in an interval) :The average velocity of a moving particle over a certain time interval is defined as thedisplacement divided by the lapsed time.

Average Velocity = erval inttime

ntdisplaceme

for straight line motion, along xaxis, we have

vav = v = <v> = t

x

=

if

if

tt

xx

The dimension of velocity is [LT-1] and its SI unit is m/s.The average velocity is a vector in the direction of displacement. For motion in a straightline, directional aspect of a vector can be taken care of by +ve and -ve sign of thequantity.

2.5 Instantaneous Velocity (at an instant) :The velocity at a particular instant of time is known as instantaneous velocity. The term�velocity� usually means instantaneous velocity.

Vinst.

=

t

xlim

0t =

dt

dx

In other words, the instantaneous velocity at a given moment (say , t) is the limitingvalue of the average velocity as we let t approach zero. The limit as t 0 is writtenin calculus notation as dx/dt and is called the derivative of x with respect to t.

Note :

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.


Example 2. A particle starts from a point A and travels along the solid curve shown in figure. Find approximatelythe position B of the particle such that the average velocity between the positions A and B has thesame direction as the instantaneous velocity at B.

Answer : x = 5m, y = 3mSolution : The given curve shows the path of the particle starting at y = 4 m.

Average velocity = takentimentdisplaceme

; where displacement is straight line distance between points

Instantaneous velocity at any point is the tangent drawn to the curve at that point.

Now, as shown in the graph, line AB shows displacement as well as the tangent to the given curve.Hence, point B is the point at which direction of AB shows average as well as instantaneous velocity.

2.6 Average Speed (in an interval)

Average speed is defined as the total path length travelled divided by the total timeinterval during which the motion has taken place. It helps in describing the motion alongthe actual path.

Average Speed = interval time

travelled distance

The dimension of velocity is [LT-1] and its SI unit is m/s.

Note :

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.


Example 3. In the example 1, if Ram takes 4 seconds and Shyam takes 5 seconds to go from P to Q, find(a) Average speed of Ram and Shyam?(b) Average velocity of Ram and Shyam?

Solution : (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)

Example 4. A particle travels half of total distance with speed v1 and next half with speed v

2 along a

straight line. Find out the average speed of the particle?Solution : Let total distance travelled by the particle be 2s.

Time taken to travel first half = 1v

s

Time taken to travel next half = 2vs

Average speed = taken time Totalcovered distance Total

=

21 vs

vs

s2

= 21

21

vvvv2

(harmonic progression)

Example 5. A person travelling on a straight line moves with a uniform velocity v1 for some time and with

uniform velocity v2 for the next equal time. The average velocity v is given by

Answer :2vv

v 21 (Arithmetic progression)

Solution :

As shown, the person travels from A to B through a distance S, where first part S1 is travelled

in time t/2 and next S2 also in time t/2.

So, according to the condition : v1 =

2/tS1 and v

2 =

2/tS2

Average velocity = taken time Total

ntdisplaceme Total =

tSS 21 =

t2

tv2

tv 21

= 2

vv 21

2 .6 Average accelerat ion ( in an interval) :

The average acceleration for a finite time interval is defined as :

Average acceleration = ervalinttimevelocityinchange

Average acceleration is a vector quantity whose direction is same as that of the changein velocity.

ava

= t

v

= t

vv if


Since for a straight line motion the velocities are along a line, therefore

aav

= t

v

=

if

if

tt

vv

(where one has to substitute vf and v

i with proper signs in one dimensional motion)

2.8 Instantaneous Accelerat ion (at an instant) :

The instantaneous acceleration of a particle is its acceleration at a particular instant oftime. It is defined as the derivative (rate of change) of velocity with respect to time. Weusually mean instantaneous acceleration when we say � acceleration�. For straight motionwe define instantaneous acceleration as :

a = dt

dv =

t

vlim

0t and in general a

=

dt

vd

=

t

vlim

0t

The dimension of acceleration is [LT-2] and its SI unit is m/s2.

3 . GRAPHICAL INTERPRETATION OF SOME QUANTITIES

3.1 Average Velocity

If a particle passes a point P (xi) at time t = t

i and reaches Q (x

f) at a later time instant

t = tf , its average velocity in the interval PQ is V

av =

t

x

=

if

if

tt

xx

This expression suggests that the average velocity is equalto the s lope of the l ine (chord) joining the pointscorresponding to P and Q on the xt graph.

x

Q

P

tO

xf

xi

ti tf3.2 Instantaneous Veloc ity

Consider the motion of the particle between the two points P and Q on the xt graphshown. As the point Q is brought closer and closer to the point P, the time intervalbetween PQ (t, t , t,......) get progressively smaller. The average velocity for eachtime interval is the slope of the appropriate dotted line (PQ, PQ, PQ......).

As the point Q approaches P, the time interval approacheszero, but at the same time the slope of the dotted lineapproaches that of the tangent to the curve at the point P.As t 0, V

av (=x/t) V

inst.

Geometrically, as t 0, chord PQ tangent at P.

Hence the instantaneous velocity at P is the slope of the tangentat P in the x t graph. When the slope of the x t graph ispositive, v is positive (as at the point A in figure). At C, v is negativebecause the tangent has negat ive slope. The instantaneousvelocity at point B (turning point) is zero as the slope is zero.

3.3 Instantaneous Accelerat ion :

The derivative of velocity with respect to time is the slope of thetangent in velocity time (vt) graph.

v

t

a<0

a=0

0

a>0a=0

a=0


Example 6. Position of a particle as a function of time is given as x = 5t2 + 4t + 3. Find the velocityand acceleration of the particle at t = 2 s?

Solution : Velocity; v = dt

dx = 10t + 4

At t = 2 sv = 10(2) + 4v = 24 m/s

Acceleration; a = 2

2

dt

xd = 10

Acceleration is constant, so at t = 2 s a = 10 m/s2

4 . MOTION WITH UNIFORM VELOCIT Y

Consider a particle moving along xaxis with uniform velocity u starting from the point x= x

i at t = 0.

Equations of x, v, a are : x (t) = xi + ut ; v (t) = u ; a (t) = 0

x t graph is a straight line of slope u through xi.

as velocity is constant, v t graph is a horizontal line. at graph coincides with time axis because a = 0 at all time instants.

x

O

xi

t u is negative

slope = u

v

O

u

t

positive velocity

v

O

u

t

negative velocity

5 . UNIFORMLY ACCELERATED MOTION :

If a particle is accelerated with constant acceleration in an interval of time, then the motion istermed as uniformly accelerated motion in that interval of time.For uniformly accelerated motion along a straight line (xaxis) during a time interval of t seconds,the following important results can be used.

(a)t

uva

(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 nth sec

6 . DIRECTIONS OF VECTORS IN STRAIGHT LINE MOTION

In straight line motion, all the vectors (position, displacement, velocity & acceleration) will haveonly one component (along the line of motion) and there will be only two possible directions foreach vector.

For example, if a particle is moving in a horizontal line (xaxis), the two directions areright and left. Any vector directed towards right can be represented by a positive numberand towards left can be represented by a negative number.

For vertical or inclined motion, upward direction canbe taken +ve and downward as ve

line of motion

lin

e of

mo

tion

line

of m

otio

n

+

+ +

-

- -

For objects moving vertically near the surface of the earth, the only force acting on theparticle is its weight (mg) i.e. the gravitational pull of the earth. Hence acceleration forthis type of motion will always be a = g i.e. a = 9.8 m/s2 (ve sign, because theforce and acceleration are directed downwards, If we select upward direction as positive).

Note :

If acceleration is in same direction as velocity, then speed of the particleincreases.

If acceleration is in opposite direction to the velocity then speed decreases i.e. theparticle slows down. This situation is known as retardation.

Example 7. A particle moving rectilinearly with constant acceleration is having initial velocity of10 m/s. After some time, its velocity becomes 30 m/s. Find out velocity of the particleat the mid point of its path?

Solution : 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

v2 = 102 + 2ax ____ (1)302 = v2 + 2ax ____ (2)


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

v2 = 500 v = 510 m/s

Example 8. Mr. Sharma brakes his car with constant acceleration from a velocity of 25 m/s to15 m/s over a distance of 200 m.(a) How much time elapses during this interval?(b) What is the acceleration?(c) If he has to continue braking with the same constant acceleration, how much

longer would it take for him to stop and how much additional distance would hecover?

Solution : (a) We select positive direction for our coordinate system to be the direction of thevelocity and choose the origin so that x

i = 0 when the braking begins. Then the

initial velocity is ux = +25 m/s at t = 0, and the final velocity and position are

vx = +15 m/s and x = 200 m at time t.

Since the acceleration is constant, the average velocity in the interval can befound from the average of the initial and final velocities.

vav, x

= 2

1(u

x + v

x) =

2

1(15 + 25) = 20 m/s.

The average velocity can also be expressed as

vav, x

= tx

. With x = 200 m

and t = t 0, we can solve for t:

t = xav,v

Äx =

20

200 = 10 s.

(b) We can now find the acceleration using vx = u

x + a

xt

ax =

t

uv xx =

10

2551 = 1 m/s2.

The acceleration is negative, which means that the positive velocity is becomingsmaller as brakes are applied (as expected).

(c) Now with known acceleration, we can find the total time for the car to go fromvelocity u

x = 25 m/s to v

x = 0. Solving for t, we find

t = x

xx

a

uv =

1250

= 25 s.

The total distance covered is

x = xi + u

xt +

2

1a

xt2

= 0 + (25)(25) + 21

(1)(25)2

= 625 � 312.5

= 312.5 m.Additional distance covered

= 312.5 � 200

= 112.5 m.


Example 9. A police inspector in a jeep is chasing a pickpocket an a straight road. The jeep isgoing at its maximum speed v (assumed uniform). The pickpocket rides on the motor-cycle of a waiting friend when the jeep is at a distance d away, and the motorcyclestarts with a constant acceleration a. Show that the pick pocket will be caught if

2adv .

Solution : Suppose the pickpocket is caught at a time t after motorcycle starts. The distancetravelled by the motorcycle during this interval is

2at21

s ____ (1)

During this interval the jeep travels a distance

vtds ____ (2)

By (1) and (2),

vtdat2

1 2

or,a

ad2vvt

2

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

2adv2 or, 2adv

Example 10. A man is standing 40 m behind the bus. Bus starts with 1 m/sec2 constant accelerationand also at the same instant the man starts moving with constant speed 9 m/s. Findthe time taken by man to catch the bus.

x = 0t = 0

40 m t = 0x = 40

1m/sec2

Solution : Let after time �t� man will catch the bus

For 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.


Example 11. A particle is dropped from height 100 m and another particle is projected vertically upwith velocity 50 m/s from the ground along the same line. Find out the position wheretwo particle will meet? (take g = 10 m/s2)

Solution : Let the upward direction is positive.

y=0m

y=100m u=0 m/s

u=50 m/s

A

B

Let the particles meet at a distance y from the ground.For particle A,

y0 = + 100 m

u = 0 m/sa = 10 m/s2

y = 100 + 0(t) 21 10 t2 [y = y

0 + ut +

21

at2]

= 100 - 5t2 ---- (1)For particle B,

y0 = 0 m

u = + 50 m/s a = 10 m/s2

y = 50(t) 21 10 t2

= 50t 5t2 ---- (2)According to the problem;

50t 5t2 = 100 5t2

t = 2 secPutting t = 2 sec in eqn. (1),

y = 100 20 = 80 mHence, the particles will meet at a height 80 m above the ground.

Example 12. A particle is dropped from a tower. It is found that it travels 45 m in the last second ofits journey. Find out the height of the tower? (take g = 10 m/s2)

Solution :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

7 . REACTION TIME

When a situation demands our immediate action. It takes some time before we really respond.Reaction time is the time a person takes to observe, think and act.

Example 13. A stone is dropped from a balloon going up with a uniform velocity of 5 m/s. If theballoon was 60 m high when the stone was dropped, find its height when the stone hitsthe ground. Take g = 10 m/s2.


Solution : S = ut + 21

at2

� 60 = 5(t) + 21

(�10) t2

60m

+ve

�ve

� 60 = 5t � 5t2

5t2 � 5t � 60 = 0

t2� t � 12 = 0

t2 � 4t + 3t � 12 = 0

(t � 4) (t + 3) = 0

t = 4Height of balloon from ground at this instant

= 60 + 4 × 5 = 80 m

Example 14. A balloon is rising with constant acceleration 2 m/sec2. Two stones are released fromthe balloon at the interval of 2 sec. Find out the distance between the two stones 1 sec.after the release of second stone.

Solution : Acceleration of balloon = 2 m/sec2

Let at t = 0, y = 0 when the first stone is released.

By the question, y1 = 0 t

1 +

21

gt1

2

(taking vertical upward as � ve and downward as + ve)

Position of st stone = 29

g

(1 second after release of second stone will be the 3rd second for the 1st stone)

For second stone y2 = ut

2 +

21

gt2

2

u = 0 + at = � 2 × 2 = � 4m/s (taking vertical upward as � ve and downward as + ve)

y2 = � 4 × 1 +

21

g × (1)2 (t2 = 1 second)

The 2nd stone is released after 2 second

y = � 21

at2 = � 21

× 2 × 4 = � 4

So, Position of second stone from the origin = � 4 + 21

g � 4

Distance between two stones = 21

g × 9 � 21

g × 1 + 8 = 48 m.

Note :

As the particle is detached from the balloon it is having the same velocity as that ofballoon, but its acceleration is only due to gravity and is equal to g.

8 . STRAIGHT LINE-EQUATION, GRAPH, SLOPE (+ve, �ve, zero slope) .

If is the angle at which a straight line is inclined to the positive direction of xaxis, &0° < 180°, 90°, then the slope of the line, denoted by m, is defined by m = tan . If is90°, m does not exist, but the line is parallel to the yaxis. If = 0, then m = 0 & the line isparallel to the x-axis.


Slope intercept form : y = mx + c is the equation of a straight line whose slope is m &which makes an intercept c on the yaxis.

m = slope = tan = dxdy

Cx

+ve slope

y

C

x

slope = 0

y

C

x

�ve slope

y

9. PARABOLIC CURVE-EQUATION, GRAPH (VARIOUS SITUATIONS UP, DOWN, LEFT,

RIGHT WITH CONDITIONS)

x

y

y = kx2x

y

y = �kx2

x

y

x = ky2 x

y

x = �ky2

Where k is a positive constant.

Equation of parabola :

Case (i) : y = ax2 + bx + cFor a > 0

x

y

The nature of the parabola will be like that of the of nature x2 = kyMinimum value of y exists at the vertex of the parabola.

ymin

= a4D

where D = b2 � 4ac

Coordinates of vertex =

a4D

,a2b

Case (ii) : a < 0

x

y

The nature of the parabola will be like that of the nature of x2 = �ky

Maximum value of y exists at the vertex of parabola.

ymax

= a4

D where D = b2 � 4ac


10. 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 is a linear polynomial in terms of t. Hence vt 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 graph

11. INTERPRETATION OF SOME MORE GRAPHS

11.1 Position vs Time graph

( i ) Zero Veloc ity

As position of particle is fixed at all the time,so the body is at rest.

Slope;dtdx

= tan = tan 0º = 0

Velocity of particle is zero

( i i ) Uniform Velocity

Here tanis constant tan = dt

dx

dt

dx is constant.

velocity of particle is constant.

( i i i ) Non uniform velocity (increasing with time)

In this case;As time is increasing, is also increasing.

dtdx

= tan is also increasing

Hence, velocity of particle is increasing.


( i v ) Non uniform velocity (decreasing with time)

In this case;As time increases, decreases.

dtdx

= tan also decreases.

Hence, velocity of particle is decreasing.

11.2 Velocity vs t ime graph

( i ) Zero acceleration

Velocity is constant.tan = 0

dtdv

= 0

Hence, acceleration is zero.

( i i ) Uniform accelerat ion

tan is constant.

dtdv

= constant

Hence, it shows constant acceleration.

( i i i ) Uniform retardation

Since > 90º

tan is constant and negative.

dtdv

= negative constant

Hence, it shows constant retardation.

11.3 Acceleration vs time graph

(i) Constant accelerat ion

tan = 0

dtda

= 0

Hence, acceleration is constant.

( i i ) Uniformly increasing accelerat ion

is constant.

0º < < 90º tan > 0

dtda

= tan = constant > 0

Hence, acceleration is uniformly increasing with time.


( i i i ) Uniformly decreasing acceleration

Since > 90º

tan is constant and negative.

dtda

= negative constant

Hence, acceleration is uniformly decreasing with time

Example 15.The displacement vs time graph of a particle moving alonga straight line is shown in the figure. Draw velocity vs timeand acceleration vs time graph.

Solution : x = 4t2

v = dtdx

= 8t

Hence, velocity-time graph is a straight linehaving slope i.e. tan = 8.

a = dtdv

= 8

Hence, acceleration is constant throughout and is equal to 8.

Example 16. At the height of 100 m, a particle A is thrown up withV = 10 m/s, B particle is thrown down with V =10m/sand C particle released with V = 0 m/s. Draw graphs ofeach particle.(i) Displacement�time

10m

/s 0 m/sec

C�v

2

10m

/s

100

m

A

�v2

B

(ii) Speed�time

(iii) Velocity�time

(iv) Acceleration�time

Solution : For particle A :(i) Displacement vs time graph is

y = ut + 21

at2

u = + 10 m/sec2

y = 10t � 21

×10t2

510152025

�2 �1

�5

�10

�15

�20

�25

�100

1 2 3 4 5 5.5T

Dis

plac

em

ent

time = 10 t � 5t2

v = dtdy

= 10 � 10 t = 0

t = 1 ; hence, velocity is zero at t = 110 t � 5 t2 = � 100

t2 � 2t � 20 = 0

t = 5.5 sec.i.e. particle travels up till 5.5 seconds.


(ii) Speed vs time graph :Particle has constant acceleration = g throughout the motion, so v-t curve willbe straight line.when moving up, v = u + at0 = 10 � 10 t or t = 1 is the time at which speed is zero.

there after speed increases at constant rate of 10 m/s2 .

Resulting Graph is : (speed is always positive). This shows that particle travels till a time of

1 + 21 seconds

(iii) Velocity vs time graph :V = u + atV = 10 � 10t ; this shows that velocity

becomes zero at t = 1 secand thereafter the velocity is negativewith slope g.

(iv) Acceleration vs time graph :

�10

Acc

eler

atio

nTime

throughout the motion,particle has constantacceleration = �10 m/s2.

For particle B :

u = � 10 m/s. y = � 10t � 21

(10) t2 = � 10t � 5t2

(i) Displacement time graph :y = 10t � 5t2

dtdy

= � 10t � 5t2 = � 10 � 10t

this shows that slope becomesmore negative with time.

(ii) Speed time graph :

dtdy

= � 10t � 5t2 = � 10 � 10t

hence, speed is directlyproportional to time withslope of 10. initial speed = 10 m/s


(iii) Velocity time graph :

�10 m/s

Vel

ocity

Time

t

dtdy

= � 10t � 5t2 = � 10 � 10t

hence, velocity is directlyproportional to time withslope of �10. Initial velocity = �10 m/s

(iv) Acceleration vs time graph : throughout the motion,particle has constant acceleration = �10 m/s2.

a = dtdv

= � 10

For Particle C :(i) Displacement time graph :

u = 0 , y = �21

× 10t2 = � 5t2

this shows that slope becomesmore negative with time.

(ii) Speed vs time graph :

v = dtdy

= � 10 t

hence, speed is directlyproportional to time withslope of 10.

(iii) Velocity time graph :

�10

Vel

ocity

Time

20

20

V = u + atV = � 10t ;

hence, velocity is directlyproportional to time withslope of �10.


(iv) Acceleration vs time graph :

�10 m/s

Acc

eler

atio

n

Time

throughout the motion,particle has constantacceleration = �10 m/s2.

12. DISPLACEMENT FROM v - t GRAPH & CHANGE IN VELOCITY FROM a -t GRAPH

Displacement = x = area under vt graph.Since a negative velocity causes a negative displacement, areas belowthe time axis are taken negative. In similar way, can see that v = a tleads to the conclusion that area under a t graph gives the changein velocity v during that interval.

Example 17. Describe the motion shown by the following velocity-time graphs.

(a) (b)

Solution : (a) During interval AB: velocity is +ve so the particle is moving in +ve direction,but it is slowing down as acceleration (slope of v-t curve) is negative. Duringinterval BC: particle remains at rest as velocity is zero. Acceleration is alsozero. During interval CD: velocity is -ve so the particle is moving in -ve direc-tion and is speeding up as acceleration is also negative.

(b) During interval AB: particle is moving in +ve direction with constant velocityand acceleration is zero. During interval BC: particle is moving in +ve directionas velocity is +ve, but it slows down until it comes to rest as acceleration isnegative. During interval CD: velocity is -ve so the particle is moving in -vedirection and is speeding up as acceleration is also negative.

Important Points to Remember

For uniformly accelerated motion (a 0), xt graph is a parabola (opening upwards ifa > 0 and opening downwards if a < 0). The slope of tangent at any point of the parabolagives the velocity at that instant.

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

In general, the slope of tangent in xt graph is velocity and the slope of tangent in vtgraph 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 takeall areas as positive.

Area under vt graph gives displacement, if areas below the taxis are taken negative.


Example 18. For a particle moving along x-axis, velocity-time graph isas shown in f igure. Find the distance travelled anddisplacement of the particle?

Solution :Distance travelled = Area under v-t graph (taking all areas as +ve.)Distance travelled = Area of trapezium + Area of triangle

= 8622

1 + 54

2

1

= 32 + 10 = 42 m

Displacement = Area under v-t graph (taking areas below time axis as �ve.)

Displacement = Area of trapezium Area of triangle

= 8622

1 54

2

1

= 32 10= 22 m

Hence, distance travelled = 42 m and displacement = 22 m.

13. MOTION WITH NON-UNIFORM ACCELERATION

(USE OF DEFINITE INTEGRALS)

x = f

i

t

t

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

f)

The expression on the right hand side is called the definite integral of v(t) betweent = t

i and t = t

f. . Similarly change in velocity

v = vf v

i =

f

i

t

t

dt)t(a

13.1 Solving Problems which Involves Non uniform Acceleration

( i ) Acceleration depending on velocity v or time t

By definition of acceleration, we have a = dt

dv. If a is in terms of t,

v

v0

dv = t

0

dt)t(a . If a is in terms of v,

t

0

v

v

dt)v(a

dv

0

.

On integrating, we get a relation between v and t, and then

using x

x0

dx = t

0

dt)t(v , x and t can also be related.


( i i ) Acceleration depending on velocity v or posit ion x

a = dt

dv a =

dx

dv

dt

dx a =

dt

dx

dx

dv

a = vdx

dv

This is another important expression for acceleration.If a is in terms of x,

v

v0

dvv = x

x0

dx)x(a .

If a is in terms of v,

x

x

v

v 00

dx)v(a

dvv

On integrating, we get a relation between x and v.

Using x

x 0

)x(vdx

= t

0

dt , we can relate x and t.

Example 19. 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?

Solution : As acceleration is given as a function of time,

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 (

2t2

- 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 .


Example 20. For a particle moving alongv + x-axis, acceleration is given as a = x. Find the positionas a function of time? Given that at t = 0 , x = 1 v = 1.

Solution : 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

Example 21. For a particle moving along x-axis, acceleration is given as a = v. Find the position as afunction of time ?Given that at t = 0 , x = 0 v = 1.

Solution : 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

Problem 1. A particle covers 43

of total distance with speed v1 and next

41

with v2. Find the average

speed of the particle?

Answer :21

21

3vvv4v

Solution : 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

Problem 2. A car is moving with speed 60 Km/h and a bird ismoving with speed 90 km/h along the same directionas shown in figure. Find the distance travelled by thebird till the time car reaches the. tree? 240 m

Answer : 360 m


Solution : Time taken by a car to reaches the tree (t) = hr/km60m240

= hr6024.0

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

= 6024.0

90 = 0.12 × 3 km = 360 m.

Problem 3 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 are used.Find (a) the dimensions of A, B, C and D, (b) the velocity of the particle at t = 4 s, (c)the acceleration of the particle at t = 4s, (d) the average velocity during the interval t =0to t = 4s, (e) the average acceleration during the interval t = 0 to t = 4 s.

Answer : (a) [A] = [LT -3], [B] = [LT -2], [C] = [LT -1] and [D] = [L] ;(b) 78 m/s ; (c) 32 m/s2 ; (d) 30 m/s ; (e) 20 m/s2

Solution : 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

Problem 4. For a particle moving along x-axis, velocity is given as a function of time as v = 2t2 + sin t.At t = 0, particle is at origin. Find the position as a function of time?

Solution : v = 2t2 + sin t dtdx

= 2t2 + sin t

x

0

dx = dt)tsint2(t

0

2 = 1tcost

32

x 3 Ans.

Problem 5. A car decelerates from a speed of 20 m/s to rest in a distance of 100 m. What was itsacceleration, assumed constant?

Solution : v = 0 u = 20 m/s s = 100 m as v2 = u2 + 2 as0 = 400 + 2a × 100 a = � 2 m/s

acceleration = 2 m/s2 Ans.


Problem 6. A 150 m long train accelerates uniformly from rest. If the front of the train passes a railway worker50 m away from the station at a speed of 25 m/s, what will be the speed of the back part of thetrain as it passes the worker?

Solution : v2 = u2 + 2as25 × 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.

Problem 7. A particle is thrown vertically with velocity 20 m/s . Find (a) the distance travelled bythe particle in first 3 seconds, (b) displacement of the particle in 3 seconds.

Answer : 25m, 15mSolution : Highest point say B

VB = 0

v = u + gt0 = 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.

Problem 8. A car accelerates from rest at a constant rate for some time after which it deceleratesat a constant rate to come to rest. If the total time elapsed is t. Find the maximumvelocity acquired by the car.

Solution : t = t1 + t

2

slope of OA curve = tan = = 1

max

t

vvmax

O

t2t1

B

A

V

t

slope of AB curve = = 2

max

t

v

t = t1 + t

2

t =

maxv +

maxv v

max = t

Problem 9. In the above question find total distance travelled by the car in time �t� .

Solution : vmax

= t)(

t

1 =

maxv = )(

t

t

2 =

maxv= )(

t

Total distance travelled by the car in time �t�

= 21t2

1 + v

max t

2 �

22t2

1 = 2

22

)(

t21

+ 2

22

)(

t

� 2

22

)(

t21

Area under graph (directly) = )(

t21 2

=

)(2t2

Ans.


Problem 10. The displacement vs time graph of a particle moving alonga straight line is shown in the figure. Draw velocity vs timeand acceleration vs time graph.

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

Solution : (a) The equation of motion is : x = �8t2

v = dtdx

= � 16 t ; this shows that velocity is directly proportional 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)

Problem 11. Draw displacement�time and acceleration�time graph for the given velocity�time graph.

Solution : 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 slopeand a-t curve will be a straight line.Part BC : v-t curve shows zero slope i.e. constant velocity. So, s-t curve will showconstant slope and acceleration will be zero.Part CD : v-t curve shows negative slope i.e. velocity is decreasing with time or accel-eration 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:

Problem 12. For a particle moving along x-axis, following graphs are given. Find the distance trav-elled by the particle in 10 s in each case?

Solution : (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.

Problem 13. For a particle moving along x-axis, acceleration is given as 22va . If the speed of the

particle is v0 at x = 0, find speed as a function of x.

Solution : a = 2v2 or dtdv

= 2v2 ordxdv

× dtdx

= 2v2

v dxdv

= 2v2 dxdv

= 2v

v

v0v

dv =

x

0

dx2 vv0nv = x0x2

0vv

n = 2x v = v0e2x Ans.


Type (I) : Very Short Answer Type Questions : [01 Mark Each]

1. Are rest and motion absolute or relative terms ?

2. Under what condition will the distance and displacement of a moving object will have the same magnitude?

3. What will be nature of x�t graph for a uniform motion?

4. Can x-t graph be a straight line parallel to position axis ?

5. Can x-t graph be a straight line parallel to time-axis for an object which is moving ?

6. what does slope of v-t graph represent ?

7. Two balls of different masses (one lighter and other heavier) are thrown vertically upward with same initialspeed. Which one will rise to the grater height ?

8. Is it possible that velocity of an object is zero but its acceleration is none zero ? If yes, then give an example.

9. What does speedometer of a car measure ?

Type (II) : Short Answer Type Questions : [02 Marks Each]

10. Given below are some examples of motion. State in each case, if the motion is one, two or three dimensional: (a) a kite flying on a windy day (b) a speeding car on a long straight highway (c) an insect crawling on aglobe (d) a carrom coin rebounding from the side of the board (e) a planet revolving around its star.

11. A car travelling with a speed of 90 kmh�1 on a straight road is ahead of scooter travelling with a speed of 60kmh�1,calculate velocity of car wrt to scooter. solve above problem if scooter is ahead of car.

12. Show that slope of displacement-time graph is equal to the velocity of uniform motion.

13. Delhi is at a distance of 250 km from Chandigarh. A sets out from Chandigarh at a speed of 80 kmh�1 and Bsets out at the same time from Delhi at a speed of 45 kmh�1. When will they meet each other.

14. Derive the equation

x(t) = x(0) + v(0) t + 2at21

15. �The direction in which an object moves is given by the direction of velocity of the object and not by the

direction of acceleration.� Explain the above statement with some suitable example.

16. Fig. below gives x-t plot of a particle in one dimensional motion. Three different equal intervals of time areshown. In which interval is the average speed greatest and in which is it the least ? Give the sign of averagevelocity for each interval.


17. Prove that v2 � u2 = 2aS, where symbols have their usual meanings.

18. Prove that the distance travelled by a body in nth second is given by

Snth

= u + 2a

(2n � 1)

where symbols have their usual meanings.

Type (III) : Long Answer Type Questions: [04 Mark Each]

19. Derive an expression for the distance travelled by uniformly accelerated body in t seconds. Also derive anexpression for the distance travelled by the body in the nth second.

20. Deduce the following relations analytically for a uniform motion along a straight time, where the terms havetheir usual meanings :

(i) v = u + at (ii) S = ut + 21

at2 (iii) v2 � u2 = 2aS

21. The driver of a car A going at 25 ms�1 applies the brakes, decelerates uniformly, and stops in 10s. The driverof another car B going at 15 ms�1 puts less pressure on this brakes and stops in 20 s. On the same graph,plot speed-time for each of the two cars.(a) Which of the two cars travelled farther after the brakes were applied ?(b) Add a line to the graph, which shows the car B decelerating at the same rate as the car A. How longdoes it take the car B to stop at this rate of deceleration ?

22. Draw velocity-time graph of uniform motion and prove that the displacement of an object in a time interval isequal to the area under velocity-time graph in that time interval.


PART - I : SUBJECTIVE QUESTIONSSECTION (A) : DISTANCE AND DISPLACEMENT

A-1. A car starts from P and follows the path as shown in figure. Finally car stops at R. Find the distance travelled

and displacement of the car if a = 7 m, b = 8 m and r =

11 m? [Take

7

22 ]

A-2. A man moves to go 50 m due south, 40 m due west and 20 m due north to reach a field.(a) What distance does be have to walk to reach the field ?(b) What is his displacement from his house to the field?

SECTION (B) : AVERAGE SPEED AND AVERAGE VELOCITYB-1. When a person leaves his home for sightseeing by his car, the meter reads 12352 km. When he returns

home after two hours the reading is 12416 km. During journey he stay for 15 minute at midway.(a) What is the average speed of the car during this period ?(b) What is the average velocity?

B-2. A particle covers each 31

of the total distance with speed v1, v

2 and v

3 respectively. Find the average speed of

the particle ?

SECTION (C) : VELOCITY, ACCELERATION, AVERAGE ACCELERATION

C-1. The position of a body is given by x = At + 4Bt3, where A and B are constants, x is position and t is time. Find(a) acceleration as a function of time, (b) velocity and acceleration at t = 5 s.

C-2. Find the velocity as a function of time if x = At + Bt�3 , where A and B are constants, x is position and t is time.

C-3. An athelete takes 2s to reach his maximum speed of 18 km/h after starting from rest. What is the magnitudeof his average accleration?

SECTION (D) : EQUATIONS OF MOTION AND MOTION UNDER GRAVITY

D-1. A car accelerates from 36 km/h to 90 km/h in 5 s on a straight rod. What was its acceleration in m/s2 andhow far did it travel in this time? Assume constant acceleration and direction of motion remains constant.

D-2. A train starts from rest and moves with a constant acceleration of 2.0 m/s2 for half a minute. The brakes arethen applied and the train comes to rest in one minute after applying breaks. Find (a) the total distancemoved by the train, (b) the maximum speed attained by the train and (c) the position(s) of the train at half themaximum speed. (Assume retardation to be constant)

D-3. A particle moving along a straight line with constant acceleration is having initial and final velocity as 5 m/sand 15 m/s respectively in a time interval of 5 s. Find the distance travelled by the particle and the accelera-tion of the particle. If the particle continues with same acceleration, find the distance covered by the particlein the 8th second of its motion. (direction of motion remains same)

D-4. A car travelling at 72 km/h decelerates uniformly at 2 m/s2. Calculate (a) the distance it goes before it stops,(b) the time it takes to stop, and (c) the distance it travels during the first and third seconds.


D-5. A ball is dropped from a tower. In the last second of its motion it travels a distance of 15 m. Find the heightof the tower. [take g = 10m/sec2]

D-6. A toy plane P starts flying from point A along a straight horizontalline 20 m above ground level starting with zero initial velocityand acceleration 2 m/s2 as shown. At the same instant, a manP throws a ball vertically upwards with initial velocity 'u'. Balltouches (coming to rest) the base of the plane at point B ofplane's journey when it is vertically above the man. 's' is thedistance of point B from point A. Just after the contact of ballwith 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)

SECTION (E) : GRAPH RELATED QUESTIONS

E-1. For a particle moving along x-axis, velocity-time graph is as shown in figure. Find the distance travelled anddisplacement of the particle? Also find the average velocity of the particle in intervel 0 to 5 second.

E-2. A cart started at t = 0, its acceleration varies with time as shown in figure. Find the distance travelled in 30seconds and draw the position-time graph.

E-3. Two particles A and B start from rest and move for equal time on a straight line. The particle A has anacceleration a for the first half of the total time and 2a for the second half. The particle B has an acceleration2a for the first half and a for the second half. Which particle has covered larger distance?

PART - II : OBJECTIVE QUESTIONS* Marked Questions are having more than one correct option.

SECTION (A) : DISTANCE AND DISPLACEMENT

A-1. A hall has the dimensions 10 m × 10 m × 10 m. A fly starting at one corner ends up at a farthest corner. The

magnitude of its displacement is:

(A) 5 3 m (B) 10 3 m (C) 20 3 m (D) 30 3 m

SECTION (B) : AVERAGE SPEED AND AVERAGE VELOCITYB-1. A car travels from A to B at a speed of 20 km h�1 and returns at a speed of 30 km h�1. The average speed of

the car for the whole journey is :(A) 5 km h�1 (B) 24 km h�1 (C) 25 km h�1 (D) 50 km h�1


B-2.* A person travelling on a straight line without changing direction moves with a uniform speed v1 for half

distance and next half distance he covers with uniform speed v2 . The average speed v is given by

(A) 21

21

vvvv2

v

(B) v 21 vv (C) 21 v

1v1

v2

(D) 21 v

1v1

v1

B 3. A body covers first 31

part of its journey with a velocity of 2 m/s, next 31

part with a velocity of 3 m/s

and rest of the journey with a velocity 6m/s. The average velocity of the body will be

(A) 3 m/s (B) 311

m/s (C) 38

m/s (D) 34

m/s

B 4. A car runs at constant speed on a circular track of radius 100 m taking 62.8 s on each lap. What is theaverage speed and average velocity on each complete lap? ( = 3.14)(A) velocity 10m/s, speed 10 m/s (B) velocity zero, speed 10 m/s(C) velocity zero, speed zero (D) velocity 10 m/s, speed zero

SECTION (C) : VELOCITY, ACCELERATION AND AVERAGE ACCELERATION

C 1. The displacement of a body is given by 2s = gt2 where g is a constant. The velocity of the body at any timet is:(A) gt (B) gt/2 (C) gt2/2 (D) gt3/6

SECTION (D) : EQUATIONS OF MOTION AND MOTION UNDER GRAVITY

D-1. A particle performs rectilinear motion in such a way that its initial velocity has opposite direction with itsuniform acceleration. Let x

A and x

B be the magnitude of displacements in the first 10 seconds and the next

10 seconds, then:(A) x

A < x

B(B) x

A = x

B(C) x

A > x

B

(D) the information is insufficient to decide the relation of xA with x

B.

D-2. A body starts from rest and is uniformly acclerated for 30 s. The distance travelled in the first 10 s is x1, next

10 s is x2 and the last 10 s is x

3. Then x

1 : x

2 : x

3 is the same as

(A) 1 : 2 : 4 (B) 1 : 2 : 5 (C) 1 : 3 : 5 (D) 1 : 3 : 9

D-3. A ball is dropped from the top of a building. The ball takes 0.5 s to fall past the 3 m height of a window somedistance from the top of the building. If the speed of the ball at the top and at the bottom of the window arev

T and v

B respectively, then (g = 9.8 m/sec2)

(A) vT + v

B = 12 ms�1 (B) v

T � v

B = 4.9 m s�1 (C) v

Bv

T = 1 ms�1 (D)

T

B

vv

= 1 ms�1

D-4. A stone is released from an elevator going up with an acceleration a and speed u. The acceleration andspeed of the stone just after the release is(A) a upward, zero (B) (g-a) upward, u (C) (g-a) downward, zero (D) g downward, u

D-5. The initial velocity of a particle is given by u (at t = 0) and the acceleration by f, where f = at (here t istime and a is constant). Which of the following relation is valid?

(A) v = u + at2 (B) v = u + 2

at2

(C) v = u + at (D) v = u

D-6. A stone is dropped into a well in which the level of water is h below the top of the well. If v is velocity ofsound, the time T after dropping the stone at which the splash is heard is given by

(A) T = 2h/v (B) vh

gh2

T (C) v2h

gh2

T (D) vh2

g2h

T


D-7. A student determined to test the law of gravity for himself walks off a sky scraper 320 m high with astopwatch in hand and starts his free fall (zero initial velocity). 5 second later, superman arrives at thescene and dives off the roof to save the student. What must be superman's initial velocity in order thathe catches the student just before reaching the ground ?[Assume that the superman's acceleration is that of any freely falling body.] (g = 10 m/s2)

(A) 98 m/s (B) 3

275 m/s (C)

2187

m/s (D) It is not possible

D 8. In the above question, what must be the maximum height of the skyscraper so that even supermancannot save him.(A) 65 m (B) 85 m (C) 125 m (D) 145 m

SECTION (E) : GRAPH RELATED QUESTIONS

E-1*. Figure shows position-time graph of two cars A and B. Lines are parallel.

A

0

5

x(m)

B

(A) Car A is faster than car B. (B) Car A always leads Car B.(C) Both cars are moving with same velocity. (D) Both cars have positive acceleration.

E-2. Figure shows the position time graph of a particle moving on the X-axis. (A) the particle is continuously going in positive x direction(B) area under x�t curve shows the displacement of particle

(C) the velocity increases up to a time to, and then becomes constant.

(D) the particle moves at a constant velocity up to a time to, and then stops.

E-3. In the displacement�time graph of a moving particle is shown,

the instantaneous velocity of the particle is negative at the point :

t

x

C

D

E F(A) C (B) D

(C) E (D) F

E-4. The variation of velocity of a particle moving along a straight line is shown in the figure. The distance travelledby the particle in 4 s is :

(A) 25 m (B) 30 m (C) 55 m (D) 60 mE-5. A particle starts from rest and moves along a straight line with constant acceleration. The variation of velocity

v with displacement S is :

(A)

S

v

(B)

v

S

C)

v

S

(D)

v

S


E-6. The displacement time graphs of two particles A and B are straight lines making angles of respectively 300

and 600 with the time axis. If the velocity of A is vA and that of B is v

B, then the value of

B

A

vv

is

(A) 21

(B) 3

1(C) 3 (D)

31

PART - III : ASSERTION / REASON1. STATEMENT-1: A particle having negative acceleration will slow down.

STATEMENT-2 : Direction of the acceleration is not dependent upon direction of the velocity.1. (A) STATEMENT-1 is true, STATEMENT-2 is true and STATEMENT-2 is correct explanation for

STATEMENT-1(B) STATEMENT-1 is true, STATEMENT-2 is true and STATEMENT-2 is not correct explanation for

STATEMENT-1(C) STATEMENT-1 is true, STATEMENT-2 is false(D) STATEMENT-1 is false, STATEMENT-2 is true(E) Both STATEMENTS are false

2. STATEMENT-1 : Magnitude of average velocity is equal to average speed.STATEMENT-2 : Magnitude of instantaneous velocity is equal to instantaneous speed.(A) STATEMENT-1 is true, STATEMENT-2 is true and STATEMENT-2 is correct explanation for



3. STATEMENT-1 : When velocity of a particle is zero then acceleration of particle must be zero at thatinstant.

STATEMENT-2 : Acceleration is equal to

dxdv

va , where v is the velocity at that instant..

(A) STATEMENT-1 is true, STATEMENT-2 is true and STATEMENT-2 is correct explanation forSTATEMENT-1

(B) STATEMENT-1 is true, STATEMENT-2 is true and STATEMENT-2 is not correct explanation forSTATEMENT-1

(C) STATEMENT-1 is true, STATEMENT-2 is false(D) STATEMENT-1 is false, STATEMENT-2 is true(E) Both STATEMENTS are false

4. STATEMENT-1 : A particle moves in a straight line with constant accleration. The average velocity ofthis particle cannot be zero in any time intervalSTATEMENT-2 : For a particle moving in straight line with constant acceleration, the average velocity

in a time interval is 2

vu , where u and v are initial and final velocity of the particle in the given time

interval.(A) STATEMENT-1 is true, STATEMENT-2 is true and STATEMENT-2 is correct explanation for




PART - I : SUBJECTIVE QUESTIONS1. Figure shows four paths along which objects move from a starting point to a final point (particle is

moving along the same straight line), all in the same time. The paths pass over a grid of equally spacedstraight lines. Rank the paths according to

(a) the magnitude of average velocity of the objects(b) the average speed of the objects, greatest first.

2. A particle moving in straight line, traversed half the distance with a velocity v0. The remaining part of the

distance was covered with velocity v1 for half the time and with velocity v

2 for the other half of the time.

Find the mean velocity of the particle averaged over the whole time of motion.

3. The displacement of a particle moving on a straight line is given by x = 16t � 2t2. Find out(a) Displacement upto 2 and 6 s.(b) Distance travelled upto 2 and 6 s.

4. A man walking with a speed ' v

' constant in magnitude and direction passes under a lantern hanging at

a height H above the ground (consider lantern as a point source). Find the velocity with which the edgeof the shadow of the man's head moves over the ground, if his height is '

h

'.

5. A police jeep is chasing a culprit going on a moter bike. The motor bike crosses a turn at a speed of 72 km/h. The jeep follows it at a speed of 108 km/h, crossing the turn 10 seconds later than bike (keeping constantspeed). After crossing the turn, jeep acclerates with constant accleration 2 m/s2. Assuming bike travels atconstant speed, how far from the turn will the jeep catch the bike?

6. A healthy youngman standing at a distance of 6 m from a 11.5 m high building sees a kid slipping from thetop floor. With what uniform acceleration (starting from rest) should he run to catch the kid at the arms height(1.5 m)? Take g = 10 m/s2.

7. A lift is descending with uniform acceleration. To measure the acceleration , a person in the lift drops a coinat the moment when lift was descending with speed 6 ft/s. The coin is 5 ft above the floor of the lift at time itis dropped. The person observes that the coin strikes the floor in 1 second. Calculate from these data, theacceleration of the lift. [Take g = 32 ft/s2]

8. A body starts with an initial velocity of 10 m/s and moves along a straight line with a constant acceleration.When the velocity of the particle becomes 50 m/s the acceleration is reversed in direction withoutchanging magnitude. Find the velocity of the particle when it reaches the starting point.

9. The accompanying figure shows the velocity v of a particle moving on a coordinate line.

-4

2

(m/s)

(a) When does the particle move forward? move backward? Speed up? slow down?(b) When is the particle's acceleration positive? Negative ? zero?(c) When does the particle move at its greatest speed ?(d) When does the particle stand still for more than an instant?


10. A point moves rectilinearly in one direction. Fig. showsthe displacement s traversed by the point as a function of the time t.Using the plot find:

(a) the average velocity of the point during the time of motion;

(b) the maximum velocity;

(c) the time t0 at which the instantaneous velocity is equal to the mean

velocity averaged over the first t0 seconds.

11. A lift starts from the top of a mine shaft and descends with a constant speed of 10 m/s.4 s later a boy throws a stone vertically upwards from the top of the shaft with a speed of30 m/s. Find when and where stone hits the lift.[ Take: g = 10 m/s² ]

PART - II : OBJECTIVE QUESTIONSSingle choice type1. Two balls of equal masses are thrown upward, along the same vertical line at an interval of 2 seconds,

with the same initial velocity of 40 m/s. Then these collide at a height of (Take g = 10 m/s2)(A) 120 m (B) 75 m (C) 200 m (D) 45 m

2. A body starts from the origin and moves along the X-axis such that the velocity at any instant is givenby (4t3 � 2t), where t is in second and velocity is in m/s. What is acceleration of the particle, when it is(A) 28 m/s2 (B) 22 m/s2 (C) 12 m/s2 (D) 10 m/s2

3. The displacement (from origin) of a body in motion is given by x = a sin (t + ). The time at which thedisplacement is maximum is ( a, and are constants)

(A)

(B)

2 (C) / 2 (D)

2

4. A body is released from the top of a tower of height h metre. It takes T seconds to reach the ground.Where is the ball at the time T/2 seconds ?(A) at h/4 metre from the ground (B) at h/2 metre from the ground(C) at 3h/4 metre from the ground (D) depends upon the mass of the ball

5. A stone is thrown vertically upward with an initial speed u from the top of a tower, reaches the groundwith a speed 3u. The height of the tower is:

(A) g

3u2

(B) g

4u2

(C) g

6u2

(D) g

9u2

6. A particle starts from rest with uniform acceleration a. Its velocity after n seconds is v. The displacementof the particle in the last two seconds is :

(A) n

1)-2v(n(B)

n1)-v(n

(C) n

1)v(n (D)

n1)2v(2n

7. A ball is thrown vertically upwards from the top of a tower of height h with velocity v. The ball strikes theground after time.

(A)

2v

gh211

gv

(B)

2v

gh211

gv

(C)

2/1

2v

gh21

gv

(D)

2/1

2v

gh21

gv


8. A balloon is moving upwards with velocity 10 ms�1. It releases a stone which comes down to the groundin 11 s. The height of the balloon from the ground at the moment when the stone was dropped is :(A) 495 m (B) 592 m (C) 460 m (D) 500 m

9. Water drops fall at regular intervals from a tap which is 5m above the ground. The third drop is leavingthe tap at the instant the first drop touches the ground. How far above the ground is the second drop atthat instant ? (Take g = 10 ms�2)

(A) 45 m (B) 4 m (C)

25 m (D)

415 m

10. Two particles held at different heights a and b above the ground are allowed to fall from rest. The ratioof their velocities on reaching the ground is :

(A) a : b (B) b:a (C) a2 : b2 (D) a3 : b3

11. In the one-dimensional motion of a particle, the relation between position x and time t is given byx2 + 2x = t (here x > 0). Choose the correct statement :

(A) The retardation of the particle is 3)1x(4

1

(B) The uniform acceleration of the particle is 3)1x(

1

(C) The uniform velocity of the particle is 3)1x(

1

(D) The particle has a variable acceleration of 4t + 6.

12. Mark the correct statement :(A) Ideal speedometer shows the magnitude of the instantanteous velocity.(B) The magnitude of displacement in an interval is equal to distance covered in that time interval.(C) It is possible to have a situation in which the speed of a particle is some time zero and some time non-zero but the average speed is zero.(D) It is possible to have a situation in which the speed of the particle is never zero but the average speed inan interval is zero.

13. Figure shows the position of a particle movingon X-axis as function of time.

(A) The particle has come to rest 5 times(B) Initial speed of particle was zero(C) The velocity remains positive for t = 0 to t = 6 s(D) The average velocity for the total period shown is negative.

14. A body freely falling from rest has a velocity v after it falls through distance h. The distance it has to falldown further for its velocity to become double is :(A) h (B) 2h (C) 3h (D) 4h

More than one choice type15. The acceleration time plot for a particle (starting from rest) moving

on a straight line is shown in figure. For given time interval,

(A) The particle has zero average acceleration(B) The particle has never turned around.(C) The particle has zero displacement(D) The average speed in the interval 0 to 10s is the same as the average speed in the interval 10s to 20s.


16. The acceleration of a particle is zero at t = 0(A) Its velocity must be constant.(B) The speed at t = 0 may be zero.(C) If the acceleration is zero from t = 0 to t = 5 s, the speed is constant in this interval.(D) If the speed is zero from t = 0 to t = 5 s the acceleration is also zero in the interval.

17. Consider the motion of the tip of the minute hand of a clock. In one hour(A) the displacement is zero (B) the distance covered is zero(C) the average speed is zero (D) the average velocity is zero

18. Mark the correct statements for a particle going on a straight line (x�position coordinate, v�velocity, a�

acceleration) :(A) If the v and a have opposite sign, the object is slowing down.(B) If the x and v have opposite sign, the particle is moving towards the origin.(C) If the v is zero at an instant, the a should also be zero at that instant.(D) If the v is zero for a time interval, then a is zero at every instant within the time interval.

PART - III : MATCH THE COLUMN

1. Column gives some graphs for a particle moving along x-axis in positive x�direction. The variables v, x and

t represent velocity of particle, x�coordinate of particle and time respectively. Column gives certain result-ing interpretation. Match the graphs in Column with the statements in Column .

Column Column

(A)

v

v - x graph

x(p) Acceleration of particle is uniform

(B)

v2

v - x graph2

x

(q) Acceleration of particle is nonuniform

(C)

v

v - t graph

t(r) Acceleration of particle is directly proportional to �t�

(D)

v

v - t graph2

t2

(s) Acceleration of particle is directly proportional to �x�.


2. Match the following :Column Column

(A) Rate of change of displacement (p) Magnitude of average velocity(B) Average speed is always greater than or equal to (q) Initial to final position(C) Displacement has the same direction as that of (r) Velocity(D) Motion under gravity is considered as the case of (s) Uniform acceleration

PART - IV : COMPREHENSIONComprehension # 1

Read the following write up and answer the questions based on that.The graph below gives the coordinate of a particle travelling along the X-axis as a function of time. AM is thetangent to the curve at the starting moment and BN is tangent at the end moment (1 = 2 =120°).

1. The average velocity during the first 20 seconds is(A) � 10 m/s (B) 10 m/s (C) zero (D) 20 m/s

2. The average acceleration during the first 20 seconds is(A) � 1 m/s2 (B) 1 m/s2 (C) zero (D) 2 m/s2

3. The direction ( i� or � i� ) of acceleration during the first 10 seconds is _____________ .

4. Time interval during which the motion is retarded.(A) 0 to 20sec. (B) 10 to 20sec. (C) 0 to 10sec. (D) None of these

Comprehension # 2The position of a particle is given by

x = 2 (t � t2)where t is expressed in seconds and x is in meter. Possitive direction is twords right.

5. The acceleration of the particle is(A) 0 (B) 4 m/s2 (C) �4 m/s2 (D) None of these.

6. The maximum value of position co-ordinate of particle on positive x-axis is

(A) 1 m (B) 2 m (C) 21

m (D) 4 m

7. The particle(A) never goes to negative x-axis(B) never goes to positive x-axis(C) starts from the origin then goes upto x = 1/2 in the positive x-axis then goes to left(D) final velocity of the particle is zero

8. The total distance travelled by the particle between t = 0 to t = 1 s is :

(A) 0 m (B) 1 m (C) 2 m (D) 21

m


PART - I : IIT-JEE PROBLEMS (LAST 10 YEARS)

* Marked Questions are having more than one correct option.

1. A block is moving down a smooth inclined plane starting from rest at time t = 0. Let Sn be the distance

travelled by the block in the interval t = n � 1 to t = n. The ratio 1n

n

SS

is [JEE (Scr.), 2004, 3]

(A) n2

1n2 (B)

1n21n2

(C)

1n21n2

(D)

1n2n2

2. A particle is initially at rest, It is subjected to a linear acceleration a , as shown in the figure. The maximumspeed attained by the particle is [JEE Scr. 2004; 3]

(A) 605 m/s (B) 110 m/s (C) 55 m/s (D) 550 m/s

3. The velocity displacement graph of a particle moving along a straight line is shown.

The most suitable acceleration-displacement graph will be [JEE Scr. 2005; 3]

(A) (B) (C) (D)

PART - II : AIEEE PROBLEMS (LAST 10 YEARS)

1. If a body loses half of its velocity on penetrating 3 cm in a wooden block, then how much will it penetratemore before coming to rest? [AIEEE - 2002, 4/300](1) 1 cm (2) 2 cm (3) 3 cm (4) 4 cm

2. From a building two balls A and B are thrown such that A is thrown upwards and B downwards (bothvertically with same speed). If V

A and V

B are their respective velocities on reaching the ground, then

[AIEEE - 2002, 4/300](1) v

B> v

A(2) v

A =v

B

(3) vA> v

B(4) their velocities depends on their masses

3. Speeds of two identical cars are u and 4u at a specific instant. The ratio of the respective distances at whichthe two cars are stopped at the same instant is : [AIEEE - 2002, 4/300]

(1) 1 : 1 (2) 1 : 4 (3) 1 : 8 (4) 1 : 16


4. The coordinates of a moving particle at any time t are given by x = t3 and y = t3. The speed of the particleat time t is given by : [AIEEE - 2003, 4/300]

(1) 22 (2) 3t2 22 (3) t2 22 (4) 22

5. A car moving with a speed of 50 km/hr, can be stopped by brakes after at least 6 m. if the same car is movingat a speed of 100 km/hr, the minimum stopping distance is : [AIEEE - 2003, 4/300](1) 12 m (2) 18 m (3) 24 m (4) 6 m

6. A ball is released from the top of a tower of height h metres. It takes T seconds to reach the ground. What isthe position of the ball in T/3 seconds? [AIEEE - 2004, 4/300](1) h/9 metre from the ground (2) 7h/9 metre from the ground(3) 8h/9 metre from the ground (3) 17h/9 metre from the ground

7. An automobile travelling with a speed of 60 km/h, can brake to stop within a distance of 20 m. If the car isgoing twice as fast, ie. 120 km/h, the stopping distance will be [AIEEE - 2004, 4/300](1) 20 m (2) 40 m (3) 60 m (4) 80 m

8. The relation between time t and distance x is t = ax2 + bx, where a and b are constants. The acceleration is:[AIEEE 2005, 4.300]

(1) �2abv2 (2) 2bv2 (3) �2av3 (4) 2av3

9. A car, starting from rest, accelerates at the rate f through a distance S, then continues at constant speed for

time t and then decelerates at the rate 2f

to come to rest. If the total distance travelled is 15 S, then :

[AIEEE 2005, 4/300]

(1) S = ft (2) S = 61

ft2 (3) S = 721

ft2 (4) S = 41

ft2

10. A particle is moving eastwards with a velocity of 5 ms�1. In 10 second the velocity changes to 5 ms�1

northwards. The average acceleration in this time is : [AIEEE 2005, 4/300]

(1) 2

1 ms�1 towards north-west (2)

21

ms�2 towards north

(3) zero (4) 21

ms�2 towards north-west.

11. A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 m/s2. Hereaches the ground with a speed of 3 m/s. At what height approximately, did he bail out?

[AIEEE 2005, 4/300](1) 91 m (2) 182 m (3) 293 m (4) 111 m

12. A particle located at x = 0 at time t = 0, starts moving along the positive x-direction with a velocity v that

varies as v = x The displacement of the particle varies with time as [AIEEE-2006, 3/180](1) t1/2 (2) t3 (3) t2 (4) t

13. The velocity of a particle is v = v0 + gt + ft2 . If its position is x = 0 at t = 0, then its displacement after unit time

(t = 1) is [AIEEE 2007, 3/120]

(1) v0 + 2g + 3f (2) v

0 +

2g

+ 3f

(3) v0 + g + f (4) v

0 +

2g

+ f

14. An object moving with a speed of 6.25 m/s, is decelerated at a rate given by :

5.2dtd

where is the instantaneous speed. The time taken by the object, to come to rest, would be :[AIEEE - 2011, 4/120, �1]

(1) 1 s (2) 2 s (3) 4 s (4) 8 s


15. A particle of mass m is at rest at the origin at time t = 0. It is subjected to a force F(t) = F0e�bt in the x

direction. Its speed v(t) is depicted by which of the following curves ? [AIEEE 2012 ; 4/120, �1]

Hint : Acceleration = MassForce

(1) (2)

F bmb

0

v(t)

t

(3) (4)

BOARD LEVEL EXERCISE : ANSWER & SOLUTIONS

1. Both rest and motion are relative terms.

2. When the object moves along a straight line and

always in the same direction.

3. It will be a straight line, inclined with time-axis.

4. No. It is because, the x-t graph parallel to position

axis indicates that the position of the object is

changing at a given instant of time.

5. No. It is because, the x-t graph parallel to time-

axis indicates that the object is at rest.

6. Acceleration .

7. Both the balls will rise to the same height. It is

because, for a body moving with given initial veloc-

ity and acceleration, the distance covered by the

body does not depend on the mass of the body.

8. if an object is thrown verticle upward than at top

point velocity will be zero but acceleration will be

non zero.

9. Instantaneous speed.

10. (a) Three dimensional (b) One dimensional

(c) Three dimensional (d) Two dimensional.

(e) Two dimensional

11. Let CV

and SV

be the velocities of the car and the

scooter respectively.

Here CV

= 90 kmh�1 and SV

= 60 kmh�1

The relative velocity of the car w.r.t. the scooter,

SCCS VVV

or VCS

= VC � V

S

When the car is ahead of the scooter :

VCS

= 90 � 60 = 30 km h�1 (away from the scooter)

When the scooter is ahead of the car :

vCS

= 90 � 60 = 30 km h�1 (towards the scooter)

13. two hours

15. When an object is thrown up, the direction of mo-

tion of the object and its velocity are along verticaly

downward and upward direction but acceleration

is along verticaly downward direction. Therefore,

the direction of motion of the object is that of the

velocity and not that of acceleration.

16. The average speed in a small time interval is equal

to the slope of the x-t graph in that interval. Since

the slope of the graph is maximum in the time in-

terval 3 and least in the interval 2; the average speed

is greatest in interval 3 and is least in interval 2.


Further, the slope of x-t graph is positive in the

time intervals 1 and 2 and it is negative in the inter-

val 3. Therefore, in interval 1 and 2, the average

velocity is positive; and in the interval 3, the aver-

age velocity is negative.

18. 2n n.a.

21

n.us 21n 1na.21

)1n(us

1n22a

usss 1nnnth

21. For a uniformly accelerated motion, the speed-time

graph is a straight line, whose slope is equal to

the acceleration of the motion. Since both the cars

start with a given speed at zero time and have zero

speed at a later time, the v-t graph in each case

can be obtained by joining these points with a

straight line. Thus, AB represents v-t graph for the

car A and CD represents that for the car B.

5

10

15

20

25

2O 4 6 8 10 12 14 16 18 20

Spe

ed(m

s)

�1

C

D' B

CarB

D

A

Time (s)

CarA

(a) The distance travelled by each car is equal to

the area under its v-t graph. Therefore,

distance travelled by the car A,

x1 = area of OAB

= 21

AO × OB = 21

× 25 × 10 = 125 m

distance travelled by the car B,

x2 = area of COD

= 21

CO × OD = 21

× 15 × 20 = 150 m

Since x2 > x

1, the car B travels farther.

(b) If the car B decelerates at the same rate as the

car A, then its v-t graph should be a straight line

drawn from the point C and parallel to AB. There-

fore, dotted line CD� represents the v-t graph of the

car B decelerating at the same rate as the car A.

Since corresponding to point D�, t = 6s, the car B

will then stop in 6s.

EXERCISE # 1PART - I

SECTION (A) :

A-1. Distance travelled by the car = 48 m,Displacement of the car = 36 m

A-2. (a) 110 m (b) 50 m,tan�1

3

4 west of south

SECTION (B) :

B-1. (a) 32 km/h (b) zero B-2.313221

321

vvvvvv

vv3v

SECTION (C) :

C-1. (a) 24 Bt ; (b) A + 300 B, 120 B

C-2. A � 3 Bt�4 C-3. 25

= 2.5 m/s2

SECTION (D) :

D-1. a = 3 m/s2 ; 2

175 = 87.5 m

D-2. (a) 2700 m = 2.7 km , (b) 60 m/s,(c) 225 m and 2.25 km

D-3. 50m ; 2m/s2 ; 20 mD-4. (a) 100 m ; (b) 10 s ; (c) 19 m, 15 mD-5. 20m

D-6. (i) 20 m/s (ii) 4 m (iii) 656 m.

SECTION (E) :

E-1. distance travelled = 10 m; displacement = 6 m;

average velocity = 56

= 1.2 m/s

E-2. 2000 m. ,

,

straight line

parabolic curve

parabolic curve

x(m)

2000

1500

500

E-3. Particle B


PART - II

SECTION (A)

A-1. (B)

SECTION (B)

B-1. (B) B-2. (A,C) B 3. (A) B 4. (B)

SECTION (C)

C 1. (A)

SECTION (D) :

D-1. (D) D-2. (C) D-3. (A) D-4. (D)

D-5. (B) D-6. (B) D-7. (B) D 8. (C)

SECTION (E) :

E-1. (B,C) E-2. (D) E-3. (C) E-4. (C)

E-5. (B) E-6. (D)

PART - III

1. (D) 2. (D) 3. (D) 4. (D)

EXERCISE # 2

PART - I

1. (a) 1v

= 2v

= 3v

= 4v

(b) v4, avg

> v1, avg

= v2, avg

> v3, avg

2.

210

21vvv2

vv0v2

3. (a) 24 m, 24 m (b) 24 m, 40 m

4.

hHH

v 5. 400 m

6. 6 m/s2 7. 22 ft/s2 8. 70 m/s

9. (a) [0,1) s & (5,7)s

(1, 5)s

(1, 2) s & (5, 6) s

(0, 1) s & (3, 5) s & (6, 7) s

(b) (3, 6) s

(0, 2) s & (6, 7) s

(2, 3) s & (7, 9) s

(c) 0 s & (2, 3) s

(d) [7, 9] s

10. (a) 10 cm/s; (b) 25 cm/s; (c) t0 = 16s

11. )628( sec after the lift starts descending,

129m below the top of the shaft

PART - II1. (B) 2. (B) 3. (B) 4. (C)

5. (B) 6. (A) 7. (A) 8. (A)

9. (D) 10. (B) 11. (A) 12. (A)

13. (A) 14. (C) 15. (A,B,D)

16. (B,C,D) 17. (A,D) 18. (A,B,D)

PART - III :

1. (A) q, s (B) p (C) p (D) q, r

2. (A) r, (B) p, (C) q, (D) s

PART - IV :

1. (A) 2. (C) 3. i� 4. (C)

5. (C) 6. (C) 7. (C) 8. (B)

EXERCISE # 3

PART - I1. (B) 2. (C) 3. (B)

PART - II1. (1) 2. (2) 3. (2) 4. (2)

5. (3) 6. (3) 7. (4) 8. (3)

9. (3) 10. (1) 11. (3) 12. (3)

13. (2) 14. (2) 15. (3)

RECTILINER MOTION – JEE (MAIN+ADVANCED)

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