topic motion in 1-d · 1) a body moves along the sides ab, bc and cd of a square of side 10 meter...

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Motion In 1-D

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MOTION IN ONE DIMENSION DISTANCE AND DISPLACEMENT Distance is the actual length of the path. Distance between A and B depends on the path taken. It is

a scalar quantity. Its units are cm in CGS and m in MKS eg in diagram given below distance along path 1 and path 2 are different. Its dimension are [M0L1T0]

(1)

(2)A

B

Displacement of a particle is a position vector of its final position w.r.t. initial position. Displacement = AB = (x2 – x1) î + (y2 – y1) ĵ+ (z2 – z1) k̂

It is the characteristic property of any point i.e. depends only on final and initial positions.it is independent of the path taken.

A (x1, y1, z1)

B (x2, y2, z2)

O

Distance is the actual path travelled by a moving body, while displacement is the change in the

position. Distance is scalar, while displacement is vector both having same dimensions [L] and same SI unit

metre.

The magnitude of displacement is equal to minimum possible distance so, Distance |Displacement|

For moving particle distance can never be negative or zero, while displacement can be. (Zero displacement means that body after motion has come back to initial position) i.e Distance > 0 but |Displacement| > = or < 0

In general magnitude of displacement is not equal to distance. However it can be so if the motion is along a straight line without change in direction.

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SPEED AND VELOCITY It is the distance covered by the particle in one second. It is a scalar quantity. Its Unit in M.K.S.

Meter/Second or km/sec, in C.G.S. cm/sec. Its dimension are [M0L1T–1]. There are different type of speeds

Types of speed : (a) Instantaneous Speed: It is the speed of a particle at particular instant.

Instantaneous speed =dtdS

tSlim

0t

(b) Average speed = timeTotalcetandisTotal

(c) Uniform speed : If during the entire motion speed of the body remains same, the body is said to have uniform speed.

(d) Non-uniform speed : If speed changes, the body is said to have non-uniform speed. VELOCITY : is defined as rate of change of displacement. It is a vector quantity. Its direction is

same as that of displacement. Its Unit and dimension are same as that of speed. different type of velocity are

(a) Instantaneous velocity : It is defined as the velocity at some particular instant.

Instantaneous velocity dtrd

trlim

0t

(b) Average velocity :

Average velocity = timeTotalntdisplacemeTotal

(c) Uniform velocity : A particle is said to have uniform velocity, if magnitudes as well as direction of its velocity remains same and this is possible only when the particles moves in same straight line reversing its direction.

(d) Non-uniform velocity : A particle is said to have non-uniform velocity, if either of magnitude or direction of velocity changes (or both changes).

Velocity can be positive or negative, as it is a vector but speed can never be negative as it is the magnitude of velocity

i.e. v = | v |

by definition, v = dtds , the slope of displacement versus time graph gives velocity.

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Time tD

ispl

acem

ent

O i.e. v =

dtds = tan = slope of s-t curve

v = dtds ds = vdt

From figure vdt = dA. so, dA = ds

s = dA = vdt

time t

velo

city

O

v

Area under velocity versus time graph with proper algebraic sign gives displacement while without sign gives distance.

v

0t1

I III

II t2 t3 t

e.g. From the adjoining v-t graph. The distance travelled by body in time t3 = Area I + Area II + Area III and

the displacement of body = Area II – Area III – Area I

AVERAGE SPEED AND VELOCITY The average speed of a particle for a given interval of time is defined as the ratio of distance

travelled to the time taken, while average velocity is defined as the ratio of displacement to time taken.

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Average speed = takenTimetraveledcetanDis

i.e. vav = ts

Average velocity = takenTime

ntDisplaceme

i.e trvav

IMPORTANT POINTS 1) If after motion body comes back to its initial position avv

= 0 [as r = 0], but vav > 0 and finite (as s > 0)

2) For a moving body average speed can never be negative or zero (unless t ), while average velocity can be i.e. vav > 0 while avv

> = or < 0

3) In general average speed is not equal to magnitude of average velocity (as s | r |). However it can be so if the motion is along a straight line without change in direction (as s = | r |). 4) If a graph is plotted between distance (or displacement) and time, the slope of chord during a given time interval gives average speed (or velocity)

Dis

plac

emen

t

t

B

A

t1 t2 Time

ss2

s1

vav = ts

= tan = slope of chord

SPECIAL CASES: Case I: An Particle covers distance S1 and S2 in equal interval of time i.e t in a straight line. Total distance covered/displacement = S1 + S2 Total time taken = 2t

Average speed = 1 2S S2t

Case II: A particles moves with velocity V1 & V2 for t1 & t2 time respectively. Total distance covered = V1t1 + V2t2 Total time taken = t1 + t2

Average speed = 1 1 2 21 2

V t V tt t

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Case III: A particle covers different distance i.e S1, S2, S3 with velocity V1, V2 & V3

1 2 3avg31 2

1 2 3

S S SV

SS SV V V

Case IV: If particle moves for equal interval of time (t) with velocity V1 & V2, then 1 2avg

V t V tV

2t

1 2V V

2

SOLVED EXAMPLES 1) A body moves along the sides AB, BC and CD of a square of side 10 meter with velocity of

constant magnitude 3 meter/sec. Its average velocity will be- (A) 3 m/sec (B) 0.87 m/sec (C) 1.33 m/sec (D) None Sol. Average velocity of the body

= timeTotal

ntdisplacemeTotal = v/)CDBCAB(

AD

= 3/30

10 = 1 m/sec

Hence correct answer is (D). 2) A body covers half the distance with a velocity 10 m/s and remaining half with a velocity

15 m/s along a straight line. The average velocity will be- (A) 12 m/s (B) 10 m/s

(C) 5 m/s (D) 12.5 m/s

Sol. Average velocity v = 21

21vvvv2

51015102

= 12 m/sec

Hence correct answer is (A) 3) A point travelling along a straight line traverse one third the distance with a velocity v0. The

remaining part of the distance was covered with velocity v1 for half the time and with velocity v2 for the other half of the time. The mean velocity of the point averaged over the whole time of motion will be-

(A) )vvv(3

)vv(v

321

210

(B)

321

210

vvv)vv(v3

(C) 321

210

v4vv)vv(v

(D)

021

210

v4vv)vv(v3

Sol. Let s be the total distance. Let (s/3) distance be covered in time t1 while the remaining distance

(2s/3) in time t2 second.

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Now

3s = v0 t1 or t1 =

0v3s .......(1)

and

3s2 = v1

2t 2 + v2

2t2

or t2 = )vv(3s4

21 .......(2)

Average velocity =

)vv(s4

v3s

stt

s

210

21

= 021

210

v4vv)vv(v3

Hence correct answer is (D) 4) A particle moves along a circle of radius R. It starts from A and moves in anticlockwise

direction. Calculate the distance travelled by the particle (a) from A to B (b) from A to C, (c) from A to D (d) in one complete revolution. Also calculate the magnitude of displacement in each case. Fig.

Sol: (a) Distance travelled by particle from A to B 2 R R4 2

Displacement = 2 2| AB | OA OB

2 2R R 2R

(b) For the motion from A to C, distance travelled 2 R R2

Displacement = | AC | 2R

(c) For the motion from A to D, distance travelled 2 R 3 3 R4 2

Displacement = 2 2| AD | R R

= 2R (d) For one complete revolution, i.e., motion from A to A, total distance travelled = 2 πR Displacement = zero, since the final position coincides with the initial position. 5) In an open ground, a motorist follows a track that turns to his left by an angle 600 after every

200m. Starting from a given turn, find the displacement and the total path length covered at the second, fourth and seventh turn.

Sol: Let motorist start from A and follows the path ABCDEFA as shown in fig. The motorist reaches at position C at the end of second turn, at position E at the end of fourth turn and at position B at the end of seventh turn.

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(i) At the end of second turn. Displacement of motorist is; AC AB BC

Magnitude of AC

= 2 2 0(AB) (BC) 2AB BC cos 60 = 2 2(200) (200) 2 200 200 1/ 2 = 200√3 m Path length at the end of second turn = AB + BC = 2 AB = 2 × 200 = 400 m (ii) At the end of fourth turn, the displacement of motorist is AE BD

Magnitude of AE

= magnitude of BD

= 2 2 0(BC) (CD) 2(BC)(CD) cos 60 = 2 2(200) (200) 2 200 200 1 / 2 = 200√3m Path length at the end of fourth turn = AB + BC + CD + DE = 4 AB = 4 × 200 = 800m (iii) At the end of seventh turn, the displacement of motorists is AB

.

Magnitude of AB

= AB = 200 m Path length at the end of seventh turn = AB + BC + CD + DE + EF + FA + AB = 7 AB = 7 × 200 = 1400 m 6) A man walks 10m towards east and then at an angle of 300 to the north of east and walks 10 m.

Calculate the net displacement of the man. Also find the direction of net displacement. Sol: Refer to Figure below the net displacement of man.

Where OA = 10m = AB and angle between OA

and OB

is 300

Magnitude of OB

is 2 2 0| OB | (OA) (AB) 2(OA)(AB) cos 30

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10 1 1 1.732 10 3.732 = 10 1.93 19.3m Direction of OB

0

0

10sin 30 10 1/ 2tan10 10 cos 30 10 10 3 / 2

0

0

10sin 30 10 1/ 2tan10 10 cos 30 10 10 3 / 2

= tan 150 or 015 EXERCISE 1 Q.1 A car travels from place A to the place B at 20 km/hour and returns at 30 km/hour. The average

speed of the car for the whole journey is – (A) 25 km/hour (B) 24 km/hour (C) 50 km/hour (D) 5 km/hour Q.2 A car travels a distance of 2000 m. If the first half distance is covered at 40 km/hour and the

second half at velocity v and if the average velocity is 48 km/hour, then the value of v is - (A) 56 km/hour (B) 60 km/hour (C) 50 km/hour (D) 48 km/hour Q.3 A car travels the first half of the journey at 40 km/hour and the second half at 60 km/hour. The

average speed of a car is - (A) 40 km/hour (B) 48 km/hour (C) 52 km/hour (D) 60 km/hour Q.4 A car travels first 1/3 of the distance AB at 30 km/hr, next 1/3 of the distance at 40 km/hr, last

1/3 of the distance at 24 km/hr. Its average speed in km/hr for the whole journey is - (A) 40 (B) 35 (C) 30 (D) 28 Q.5 A train travels from one station to another at a speed of 40 km/hour and returns to the first

station at the speed of 60 km/hour. Calculate the average speed and average velocity of the train

(A) 48 km/hr, zero (B) 84 km/hr, 10 km/hr (C) 84 km/hr, zero (D) 48 km/hr, 10 km/hr Q.6 A passenger travels along a straight line with velocity v1 for first half time and with velocity v2 for next

half time, then the mean velocity v is given by -

(A) v = 2

vv 21 (B) v = 21vv

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(C) v = 1

2

vv (D)

21 v1

v1

v2

Answers 1.b 2.b 3.b 4.c 5.a 6.a

ACCELERATION

It is defined as the rate of change of velocity. It is a vector quantity. Its direction is same as that of change in velocity and not of the velocity.

Its dimensions are [LT–2] and SI units (m/s2) There are three ways possible in which change in velocity can occur a) only direction changes eg circular motion b) only magnitude changes eg motion under gravity c) both direction and magnitude changes eg: projectile motion

Types of acceleration : (a) Instantaneous acceleration : It is defined as the acceleration of a body at some particular instant. Instantaneous acceleration

= dtvd

tvlim

0t

(b) Average acceleration :

12

12av tt

vvtva

(c) Uniform acceleration : A body is said to have uniform acceleration if magnitude and direction of the acceleration

remains constant during particle motion. If a particle is moving with uniform acceleration, this does not necessarily imply that particle

is moving in straight line eg: parabolic motion (d) Non-uniform acceleration :

A body is said to have non-uniform acceleration, if magnitude or direction or both change during motion.

Important points: a) If acceleration is zero, velocity will be constant and motion will be uniform. b) if acceleration is constant then acceleration is uniform but motion is non-uniform and if

acceleration is not constant then both motion and acceleration are non-uniform. c) If a force F

acts on a particle of mass m then by Newton's II law a = F

/m

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d) Acceleration can be positive or negative. Positive acceleration means velocity is increasing with time while negative acceleration called retardation means velocity is decreasing with time.

e) dtsdv

so, 22

dtsd

dtsd

dtd

dtvda

. f) If velocity is given as function of position then by chain rule

dtdx.

dxdv

dtdva

dxdvva ]v

dtdxas[

g) As acceleration dtvda

the slope of velocity -time graph gives acceleration i.e.

dtvda

= tan

SOLVED EXAMPLE Ex.1 The velocity acquired by a body moving with uniform acceleration is 20 meter/second in first 2

seconds and 40 m/sec in first 4 sec. The initial velocity will be - (A) 0 m/sec (B) 40 m/sec (C) 20 m/sec (D) None

Sol. Acceleration = Time

velocityinChange

= 242040

= 10 m/sec2

From the relation, v = u + at 20 = u + 10 × 2 u = 0 m/sec Hence correct answer is (A). Ex.5 A particle moves along the x-axis in such a way that its x-coordinate varies with time

according to the equation x = 2 – 5t + 6t2. The initial velocity and acceleration will respectively be-

(A) 5 m/s, 12 m/s2 (B) –12 m/s, –5 m/s2

(C) 12, – 5 m/s2 (D) –5 m/s, 12 m/s2 Sol. x = 2 –5t + 6 t2,

v = dtdx = – 5 + 6 × 2 × t

Initial velocity at t = 0, v = – 5 m/sec

a = 22

dtxd = 12 m/sec2

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Hence correct answer is (D) Ex.6 The position of a body with respect to time is given by x = 4t3 – 6t2 + 20 t + 12. Acceleration

at t = 0 will be- (A) –12 units (B) 12 units

(C) 24 units (D) –24 units

So. a = 22

dtxd

dtdv

dtdx = 12t2 – 12t + 20

22

dtxd = 24t – 12

when t = 0, a = 0 – 12 = –12 units Hence correct answer is (A). EXERCISE 2 Q.1 A particle is moving east-wards with a velocity of 5 m/sec. In 10 seconds its velocity changes

to 5 m/sec north-wards. The average acceleration during this time is - (A)1/ 2 m/sec2 in N-W direction (B) 1/ 2 m/sec2 in the N-E direction (C) 1/2 m/sec2 in N-W direction (D) 1/2 m/sec2 towards east Q.2 A truck travelling due to North at 20 m/s turns East and travels at the same speed. The change

in its velocity is- (A) 20 2 m/s North-East (B) 20 2 m/s South-East (C) 40 2 m/s North-East (D) 20 2 m/s North-West Q.3 A truck travelling due north at 20m/s turns west and travels with the same speed. What is the

change in velocity ? (A) 40 m/s north-west (B) 20 2 m/s north-west (C) 40 m/s south-west Q.4 At an instant t , the co-ordinates of a particle are x = at2, y = bt2 and z = 0 , then its velocity at

the instant t will be

(A) t 22 ba (B) 2t 22 ba

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(C) 22 ba (D) 2t2 22 ba

Q.5 A particle is moving so that its displacement s is given as s = t 3– 6t2 + 3t + 4 meter. Its velocity at the instant when its acceleration is zero will be -

(A) 3 m/s (B) –12 m/s (C) 42 m/s (D) –9 m/s (D) 20 2 m/s south-west Q.6 The displacement y (in meters) of a body varies with time (in seconds) according to the

equation

y = – 32 t2 + 16t + 2. How long does the body come to rest ?

(A) 8 seconds (B) 10 seconds (C) 12 seconds (D) 14 seconds Q.7 The displacement x of a particle moving along a straight line is related to time by the relation

t = x + 3. The displacement of the particle when its velocity is zero is given by- (A) 0 (B) – 2 (C) 2 (D) 1

Q.8 The displacement-time relationship for a particle is given by x = a0 + a1t + a2t2. The

acceleration of the particle is - (A) a0 (B) a1 (C) a2 (D) 2a2 Answers

1.a 2.b 3.d 4.b 5.d 6.c 7.a 8.d

EQUATIONS OF MOTION: Let u = Initial velocity (at t = 0), v = Velocity of the particle after time t

a = Acceleration (uniform), s = Displacement of the particle during time 't'

1ST : Acceleration, t

uva

. From above equation tauv ....(i) 2ND : Displacement s = Average velocity × time,

t2

vus

....(ii)

From (i) and (ii) us t + (1/2) a t2 ....(iii)

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3RD From (i) and (ii) v 2 = u 2 + 2 s.a .....(iv) 4TH ns

= displacement of particle in nth second = ns

– 1ns

= { u (n) + (1/2) a n2} – {u (n –1) + (1/2)a

(n – 1)2} ns

= u + 1/2 a (2n – 1) Important Points

avgu v

S v .t t2

21S Vt at2

Special Case: Stopping Distance & Stopping Time A body moving with velocity u stops after a distance S (by applying heater). If same body in moving with nu speed then it will stop after V2 = u2 – 2as (-ve sign. retardation) When body stops V = 0

2u S

2a

2S u ∴ distance travelled will be n2s before stopping Time taken to stop will be V = u – at

u ta

T ∝ u ∴ it will take nt time to stop SOLVED EXAMPLES: Ex.1 A body travels 200 cm in the first two seconds and 220 cm in the next four second. The

velocity at the end of the seventh second from the start will be- (A) 10 cm/s (B) 5 cm/s

(C) 15 cm/s (D) 20 cm/s Sol. Let 'u' and 'a' be the initial velocity and acceleration of the body respectively. For first two

second (t = 2 sec), the distance covered is 200 cm.

Now using, s = u t + 21 a t2

we have, 200 = u (2) + (1/2) a (2)2 .....(1) After four seconds of this journey i.e., after a time t = 6 sec

the distance covered is 200 cm + 220 cm = 420 cm.

Hence 420 = u (6) + (1/2) a (6)2 .....(2) Solving equations (1) and (2),

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we get, u = 115 cm/sec and a = – 15 cm/sec2 Now velocity after 7 seconds = u + a t =115 + (–15) 7 = 10 cm/sec Hence correct answer is (A) Ex.2 An particle travels along the inside of straight hollow tube, 2.0 metre long, of a particle

accelerator. Under uniform acceleration, how long is the particle in the tube if it enters at a speed of 1000 m/s and leaves at 9000 m/s -

(A) 4 × 10–4 sec (B) 2 × 10–7 sec

(C) 40 × 10–4 sec (D) 20 × 10–7 sec Sol. Let a be the uniform acceleration of -particle. According to the given problem s = 2.0 m,

v = 9000 m/sec and u = 1000 m/sec

Using the formula v2 = u2 + 2 a s, we have

a = )0.2(2

)1000()9000(s2uv 2222

= 0.4

108 7 = 2.0 × 107 m/sec2

Ex.3 A truck starts from rest with an acceleration of 1.5 m/s2 while a car 150 m behind starts from rest with an acceleration of 2 m/s2. How long will it take before both the truck and car side by side, and how much distance is travelled by each ?

(A) 2.45 s, 500 m (truck), 650 m (car) (B) 5 s, 450 m (truck), 600 m (car) (C) 24.5 s, 450 m (truck), 600 m (car) (D) 5.3 s, 500 m (truck), 650 m (car)

Sol. Let x be the distance travelled by the truck when truck and car are side by side. The distance

travelled by the car will be (x + 150) as the car is 150 metre behind the truck. Applying the formula

s = ut + (1/2) a t2, we have, x = 1/2 × (1.5) t2 ........(1) and (x + 150) = (1/2) × (2) t2 ........(2) Here t is the common time. From equations (1) and (2)

we have, 5.1

2x150x

Solving we get, x = 450 m (truck)

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and x + 150 = 600 m (car) Substituting the value of x in eq. (1), we get

450 = 1/2 (1.5) t2

t = 6005.1

2450 = 24.5 sec.

Hence correct answer is (C) Ex.4 Two car travelling towards each other on a straight road at velocity 10 m/sec and 12 m/sec

respectively. When they are 150 m apart, both drivers apply their brakes and each car decelerates at 2 m/sec2 until it stops. How far apart will they be when they have both come to a stop?

(A) 8.9 m (B) 89 m (C) 809 m (D) 890 m Sol. Let x1 and x2 be the distance travelled by the car before they stop under deceleration. Using the

formula v2 = u2 + 2a s, we have, 0 = (10)2 – 2 × 2 × x1

and 0 = (12)2 – 2 × 2 × x2 Solving we get,

x1 = 25 metre and x2 = 36 metre, Total distance covered by the two cars x1 + x2 = 25 + 36 = 61 metre Distance between the two cars when they stop

= 150 – 61 = 89 metre Hence correct answer is (B) Ex.5 The driver of a train travelling at 115 km/hour sees on the same track 100 m in front of him a

slow train travelling in same direction at 25 km/hr. The least retardation that must be applied to the faster train to avoid a collision will be-

(A) 3.125 m/s2 (B) 31.25 m/s2

(C) 312.5 m/s2 (D) 0.3125 m/s2 Sol. The velocity of faster train with respect to slow train = (115 – 25) = 90 km/hr. The distance between two trains = 100 m. If the collision is to be avoided, the relative speed should become zero till distance 100 m is

covered.

Using the formula v2 = u2 + 2 a s, we have

0 = 2

18590

+ 2 a × 100

( 90 km/h = 6060

1090 3

= 90 ×

185 m/sec)

a = – 2

18590

2001

m/sec2

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= – 3.125 m/sec2

Retardation = – a = 3.125 m/s2

Hence correct answer is (A) Ex.6 A balloon going upward with a velocity of 12 m/sec is at a height of 65 m from the earth at any

instant. Exactly at this instant a packet drops from it. How much time will the packet take in reaching the earth? (g = 10 m/sec2)

(A) 7.5 sec (B) 10 sec (C) 5 sec (D) None

Sol. h = –ut + 1/2 gt2 65 = –12t + 5t2 5t2 – 12t – 65 – 0 t = 5 sec Hence correct answer is (C) EXERCISE 3 Q.1 A particle moves from the position of rest and attains a velocity of 30 m/sec after 10sec. The

acceleration will be (A) 9 m/sec2 (B) 18 m/sec2 (C) 3 m/sec2 (D) 4 m/sec2 Q.2 A particle, after starting from rest , experiences, constant acceleration for 20 seconds. If it covers

a distance of S1, in first 10 seconds and distance S2 in next 10 sec, then (A) S2 = S1/2 (B) S2 = S1 (C) S2 = 2S1 (D) S2 = 3S1 Q.3 If u is the initial velocity of a body and a the acceleration , the value of distance travelled by

nth second is :

(A) u + 21 a (2n +1) (B) u +

21 a (2n–1)

(C) u – 21 a (2n + 1) (D) u –

21 a(2n–1)

Q.4 A body starts from rest, the ratio of distances travelled by the body during 3rd and 4th seconds is : (A) 7/5 (B) 5/7 (C) 7/3 (D) 3/7

Q.5 A body starting from rest and has uniform acceleration 8 m/s2. The distance travelled by it in 5th second will be

(A) 36m (B) 40m (C) 100m (D) 200m ANSWERS

1.c 2.d 3.b 4.b 5.a

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Various Graphs Related to Motion : A. Displacement-Time Graph : (a) stationary body

TimeD

ispl

acem

ent

0

y

x

(b) body moving with constant velocity

Time

Dis

plac

emen

t

0

y

x

(c) body moving with non-uniform velocity

Time

Dis

plac

emen

t

0

y

x

(d) body with accelerated motion

Time

Dis

plac

emen

t

0

y

x

(e) For a body with decelerated motion

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Time

Dis

plac

emen

t

0

y

x

(f) For a body which returns towards the point of reference

TimeDis

plac

emen

t

0

y

x > 90º

(g) For a body whose velocity constantly changes

Time

Dis

plac

emen

t

0

y

x

B. Velocity-Time Graph: (a) For the body having constant velocity or zero acceleration

Time

Vel

ocity

0

y

x

(b) When the body is moving with constant retardation and its initial velocity is not zero

Time

Vel

ocity

y

x 0

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(c) When body moves with non-uniform acceleration and its initial velocity is zero.

Time

Vel

ocity

0

y

x

C. Acceleration-Time Graph : (a) When acceleration is constant

Time

Acc

eler

atio

n

0

y

x

(b) When initial acceleration is zero and rate of change of acceleration is non-uniform

Time

Acc

eler

atio

n

0

y

x

SOLVED EXAMPLES Ex:1 Figure below shows the distance (S) – time (t) graphs of two trains, which start moving simultaneously in the same direction. From the graphs, find:

(a) How much B is ahead of A when motion starts (b) What is the speed of B? (c) When and where A will catch B? (d) What is the difference in speeds of A and B? Sol: (a) When motion starts, B is ahead of A by the distance OC = 80 km

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(b) speed of B = 1DE 120 80 20km hCE 2 0

(c) Since the two graphs meet at point D, so A will catch B after 2 hours and at a distance of 120 km from the origin.

(d) Speed of A = 1DF 120 0 60km hOF 2 0

Difference in speed = 60 – 20 = 40 km h-1 Ex:2 The velocity time graph for linear motion is shown in fig below. Find (a) the distance travelled

and (b) displacement, between 5 and 40s. (c) Acceleration between 15 to 25 seconds.

Sol: (a) Distance travelled by body in time 5 to 15 seconds = Area A1 ABB1 =

(2 4) 10 30m2

Distance travelled by body in time 15 to 25 seconds = Area BB1C = 4 10 20m

2

Distance travelled by body in time 25 to 30 seconds = Area CDD1 = 5 2 5m

2

Distance travelled by body in time 30 to 40 seconds = Area D1 DEF = (2 4) 10 30m

2

∴ Total distance travelled between 5 to 40 seconds = 30 + 20 + 5 + 30 = 85 m (b) Here, the area CDD1 and area D1DEF being below the time axis will show negative displacement. Hence, total displacement between 5 to 40 seconds = 30 + 20 – 5 – 30 = 15 m (c) Acceleration of the body in time interval 15s to 25s is the slope of the line BC

211

BB 4 4 0.4m / sCB (25 15) 10

Ex:3 Velocity (v) time (t) graph of a car starting from rest is shown in fig. Draw acceleration (a) – time (t) graph for the motion of the car and find total distance travelled by car.

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Sol: From the graph we note that car starting from rest accelerates for time 0 to 5 s and attains a

velocity 20 ms-1. It then moves with a constant velocity 20 ms-1 from 5 s to 10 s. After that the motion of car is retarded for time 10s to 14s and it comes to rest at time t = 14s.

For time t = 0 to t = 5s, the acceleration of car = slope of v – t graph = 220 0 4ms5

For time t = 5s to t = 10s, velocity of car is constant, hence acceleration is zero. For time t = 10 s to t = 14 s.

Acceleration of car = slope of v – t graph or 20 20a 5ms4

Thus acceleration (a) – time (t) graph is of the type as shown in fig. Total distance travelled by car in 14 seconds = Are OABC = area of ∆OAE + Area of ABDE + Area ∆BCD

= 20 5 20 420 52 2

= 50 + 100 + 40 = 190 m

Ex:4 A body starting from rest accelerates uniformly along a st. line, at the rate of 10 ms-2 for 5s. It

moves for 2s with uniform velocity of 50 ms-1. Then, it retards uniformly and comes to rest in 3s. Draw velocity – time graph of the body and find the total distance travelled by body.

Sol: Taking the motion of the body from 0 to 5 seconds we have u = 0; v = ? a = 10 ms2 ; t = 5s. As, v = u + at ∴ v = 0 + 10 × 5 = 50 m/s The velocity – time graph of the motion of the body is as shown in fig.

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The distance travelled by body during motion = area of trapezium OABC

11 (OC AB) AA2

= 1 (10 2) 50 300m2

Ex:5 A ball is thrown upward with an initial velocity of 80 ms-1. After how much time will it return to ground? Draw velocity time graph for the ball and find from the graph

(a) the maximum height attained by the ball and (b) height of the ball after 12s. Take g = 10 ms-2 Sol: Here, u = 80 ms-1; At the highest point v = 0, a = -10 ms-2. Let the ball reach the highest point in time t. As v = u + at; so 0 = 80 + (-10)t or t = 80/10 = 8s The ball will return to ground in time = 2t = 2 × 8 = 16 s Velocity of the ball at different instants of time will be as follows: At t = 0, v = 80 – 10 × 0 = 80 ms-1 At t = 4, v = 80 – 10 × 4 = 80 ms-1 At t = 8, v = 80 – 10 × 8 = 0 At t = 12, v = 80 – 10 × 12 = -40 ms-1 At t = 16, v = 80 – 10 × 16 = - 80 ms-1 The velocity time graph will be a st. line ABC as shown in figure.

(a) Maximum height attained by the ball = area OAB

= 1 8 80 320m2

(b) Height attained after 12s.

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= area OAB + Area BCD

1320 (12 8) ( 40) 320 80 240m2

EXERCISE Q.1 The variation of velocity of a particle moving along straight line is shown in figure. The

distance traversed by the body in 4 seconds is –

20

Vel

ocity

(m/s

ec.)

10

1 2 3 4Time (sec)

(A) 70 m (B) 60 m (C) 40 m (D) 55 m Q.2 The v-t graph of a linear motion is shown in adjoining figure. The distance travelled in 8

seconds is –

4

m/s

ec

–2

1 2 3 45 6 7 8

(sec)

(A) 18 meters (B) 16 meters (C) 8 meters (D) 6 meters Q.3 Which one of the following curves do not represent motion in one dimension-

(A)

v

t

(B)

v

t

(C)

v

t

(D)

v

t

Q.4 The following figures show some velocity v versus time t curves. But only some of these can be realised in practice. This are-

(a) v

0 t

(b) v

0 t

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(c) v

0 t

(d)

v

0 t

(A) only a, b and d (B) only a , b , c (C) only b and c (D) all of them Q.5 The velocity-time graph of a linear motion is shown below. The displacement from the origin

after 8 seconds is -

v m

/sec

–4

1 2 3 45 6 7 8

t(sec) –2

0

2

4

(A) 18m (B) 16 m (C) 6m (D) 6 cm Q.6 The following shows the time-velocity graph for a moving object. The maximum acceleration

will be-

40

m/s

ec

20 30 40 70

60

20

t (sec)

D

C

A B

(A) 1 m/sec2 (B) 2m/sec2

(C) 3m/sec2 (D) 4m/sec2

ANSWERS

1.d 2.a 3.b 4.a 5.c 6.d

MOTION UNDER GRAVITY

In case of motion under gravity unless stated it is taken for granted that. (i) The acceleration is constant, i.e.

a = g = 9.8 m/s2 and directed vertically downwards.

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(ii) air has no effect on the motion. a) Body Falling Freely Under Gravity :

Taking initial position as origin and direction of motion (i.e. downward direction) as positive, here we have u = 0 (as body starts from rest)

a = + g (as acceleration is in the direction of motion) So, if the body acquires velocity v after falling a distance h in time t, equations of motion viz v = u + at

s = ut +

21 at2

and v 2 = u2 + 2as reduces to v = gt ....(1)

h =

21 gt2 ....(2)

v 2 = 2 gh ....(3) important cases : 1. If the body is dropped from a height H, as in time t, it has fallen a distance h from its initial

position, the height of the body from the ground will be h' = H – h, with h = 1/2 gt2.

2. As h = (1/2) gt2 i.e. h t2, distance fallen in time t, 2t, 3t etc. will be in the ratio of 12 : 22 : 32 : ––––––– i.e. square of integers.

3. The distance fallen in nth sec., hn – hn –1 = (1/2) g(n)2 – (1/2) g(n –1)2

= (1/2) g(2n –1) So distance fallen in Ist, IInd, IIIrd sec will be in the ratio 1 : 3 : 5 i.e. odd integers only.

b) Body is projected vertically up : Taking initial position as origin and direction of motion (i.e. vertically up) as positive, here we

have v = 0 [as at the highest point, velocity = 0], a = – g [as acceleration is downwards while motion upwards].

So, if the body is projected with velocity u and reaches the highest point at a distance h above the ground in time t, the equations of motion viz

v = u + at

s = ut +

21 at2

and v2 = u2 + 2as reduces to 0 = u – gt

h = ut – 1/2 gt2

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and 0 = u2 – 2gh or u = gt ....(1)

h = 1/2 gt2 ...(2)

( u = gt), u2 = 2 gh ....(3)

These equations can be used to solve most of the problems of bodies projected vertically up as, if

Important points . (a) The motion is independent to the mass of the body as in any equation of motion mass is not

involved. This is why a heavy and light body when released from same height reaches the ground simultaneously and with same velocity.

i.e. t = g/h2

and v = gh2

(b) As from case (b) time taken to reach a height h, T1 = g/h2

And from case (a) time taken to fall down through a distance h, T2 = g/h2

so T1 = T2 = g/h2

in case of motion under gravity time taken to go up is equal to the time taken to fall down through the same distance.

Special case: Body is thrown vertically upward constant air resistance (a) in taken into account

neta = g + a (downward) g ↓ a ↓ (same direction)

upward2ht

g a

neta = g – a (down ward) g ↓ a ↑ (Opposite direction)

decent2ht

g a

Ex.15 A body is falling from a height 'h'. It takes t1 sec to reach the ground, the time taken to reach

the half of the height will be-

(A) 2 t1 (B) 2t1

(C) 2

t1 (D) 2t1

Sol. Time of fall to ground = t1 = gh2 .

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Let the time taken to cover h/2 is t2,

then t2 = 2t

gh

g2h2 1

Hence correct answer is (C). Ex.16 A body thrown up with a velocity reaches a maximum height of 100 m. Another body with

double the mass thrown up with double the initial velocity will reach a maximum height of- (A) 400 m (B) 200 m

(C) 100 m (D) 250 m Sol. The maximum height reached by a vertically projected body is given by S = u2/2g, i.e. S u2 as 2g is constant,

= 22

21

2

1

uu

SS

2S

100 =41

)u2(u

2

2

S2 = 400 m Hence correct answer is (A) Ex.17 A ball dropped from the top of a building takes 0.5 sec to clear the window of 4.9 m height.

What is the height of building above the window? (A) 2.75 m (B) 5.0 m

(C) 5.5 m (D) 4.9 m Sol. As the ball clears the window of height 4.9 m. Now from, S = ut + 1/2 gt2 ,

We get, 4.9 = u × (1/2) + (1/2)9.8 × (1/2)2

2u = 4.9 –

49.4 = 4.9 ×

43

u = 2

39.4 m/sec

If h is the height of the building above the window. For the flight from top of the building.

h = 228.9239.439.4

g2u2

= 2.75 m

Hence correct answer is (A). Ex.18 A ball is thrown straight upward with an initial speed of 12 m/sec. After 1 sec velocity &

displacement will respectively be- (A) 2 m/sec, 7 m (B) 7m, 2 m/sec

(C) 2.20 m/s, 7.10 m (D) None Sol. v = u – gt = 12 – 9.8 ×1 v = 12 – 9.8 = 2.20 m/s h = ut– (1/2) gt2

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= 12 × 1 – (1/2) × 9.8 (1)2 h = 12 – 4.9 = 7.10 m.

Hence correct answer is (C) EXERCISE Q.1 An object is released from some height. Exactly after one second, another object is released from

the same height. The distance between the two objects exactly after 2 seconds of the release of second object will be:

(A) 4.9 m (B) 9.8 m (C) 19.6 m (D) 24.5 m Q.2 A stone is thrown vertically upwards from the top of a tower with a velocity u and it reaches

the ground with a velocity 3u. The height of the tower is (A) 3u2/g (B) 4u2/g (C) 6u2/g (D) 9u2/g Q.3 A ball is thrown from the ground with a velocity of 80 ft/sec. Then the ball will be at a height

of 96 feet above the ground after time (A) 2 and 3sec (B) only 3 sec (C) only 2sec (D) 1 and 2 sec Q.4 A body is dropped from a height h under acceleration due to gravity g. If t1 and t2 are time

intervals for its fall for first half and the second half distance, the relation between them is (A) t1 = t2 (B) t1 = 2t2 (C) t1 = 2.414 t2 (D) t1 = 4t2 Q.5 A stone is dropped from a bridge and it reaches the ground in 4 seconds The height of the bridge is: (A) 78.4m (B) 64m (C) 260m (D) 2000m Q.6 A rocket is launched form the earth surface so that it has an acceleration of 19.6m/s2. If its

engine is switched off after 5 seconds of its launch, then the maximum height attained by the rocket will be

(A) 245 m (B) 490 m (C) 980 m (D) 735 m Q.7 Two bodies of different masses ma and mb are dropped from two different heights, viz a and b.

The ratio of times taken by the two to drop through these distances is

(A) a : b (B) ab:

mm

b

a

(C) b:a (D) a2 : b2

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Answers 1.d 2.b 3.a 4.c 5.a 6.d 7.c

RELATIVE VELOCITY Relative motion means, the motion of a body with respect to another. Now if AV

and BV

are

velocities of two bodies relative to earth, the velocity of B relative to A will be given by ABBA VVV

: If two bodies are moving along the same line in same direction with velocities VA and VB relative to earth, the velocity of B relative to A will be given by VBA = VB – VA.

If it is positive the direction of VBA is that of B and if negative the direction of VBA is opposite to that of B.

: if the bodies are moving towards or away from each other, as direction of VA and VB are opposite, velocity of B relative to A will have magnitude

VBA = VB – (–VA) = VB + VA and directed towards A or away from A respectively.

: If a boy is running with velocity .relV

on a train moving with velocity V

relative to the ground. The velocity of boy relative to ground, v will be given by VvV .rel

VVv .rel

So, if the boy is running on the train in the direction of motion of train v = Vrel. + V And if the boy is running on the train in a direction opposite to the motion of train v = Vrel. – V

: In case of motion of a body A on a moving body B, the velocity of A relative to ground is the sum of two velocities if A is moving on B in the same direction and is equal to difference of two velocities if they are moving in opposite direction.

Solved example

Ex.21 A motor boat covers the distance between two spots on the river in t1 = 8 hr and t2 = 12 hr downstream and upstream respectively. The time required for the boat to cover this distance in still water will be-

(A) 6.9 hr (B) 9.6 hr (C) 69 sec (D) 96 sec

Sol. Let s be the distance between that two spots. Also assume that the velocity of the motor boat in still water is v and the velocity of flow of water is u.

Then , for downward journey, s/t1 = v + u …(1) For upward journey, s/t2 = v – u ...(2)

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Adding eq. (1) to (2), s/t1 + s/t2 = 2v

or t = )128(

1282)tt(

tt2vs

21

21

= 9.6 hr

Hence correct answer is (B) Exercise Q.1 A train is moving in the north at a speed 10 m/sec. Its length is 150 m. A parrot is flying

parallel to the train in the south with a speed of 5 m/s. The time taken by the parrot to cross the train will be :

(A) 12 sec (B) 8 sec (C) 15 sec (D) 10 sec Q.2 A thief is running away on a straight road on a jeep moving with a speed of 9 m/s. A police

man chases him on a motor cycle moving at a speed of 10 m/s. If the instantaneous separation of jeep from the motor cycle is 100 m, how long will it take for the policemen to catch the thief ?

(A) 1 second (B) 19 second (C) 90 second (D) 100 second Q.3 A horse rider is moving towards a big mirror with velocity v. The velocity of his image with

respect to him is - (A) 0 (B) 4 v (C) 2 v (D) v Q.4 A car A is going north-east at 80 km/hr and another car B is going south-east at 60 km/hr. Then

the direction of the velocity of A relative to B makes with the north an angle such that tan is -

(A) 1/7 (B) 3/4 (C) 4/3 (D) 3/5

Answers

1.d 2.d 3.c 4.a

topic motion in 1-d · 1) a body moves along the sides ab, bc and cd of a square of side 10 meter...

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