Class 11 - Physics - Motion In A Straight Line

Question1.

In which of the following examples of motion, can the body be considered approximately a point object:

(a) a railway carriage moving without jerks between two stations.

(b) a monkey sitting on top of a man cycling smoothly on a circular track.

(c) a spinning cricket ball that turns sharply on hitting the ground.

(d) a tumbling beaker that has slipped off the edge of a table.

When the size of an object is very small compared to scale of observation, the object can be considered as point object.

• The size railway carriage is very small compared to the distance between the two stations.
• Therefore, the carriage is considered as a point object.
• The size of monkey is very small compared to the distance covered by the cyclist. Therefore, the monkey is considered as a point object.
• The size of spinning cricket ball is very small compared to the distance through which the ball may turn as it hits the ground. Therefore, the monkey is considered as a point object.
• The height of the table is not much large than size of the beaker so it cannot be considered as a point object.

Question2.

The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in Fig. 3.19.

Choose the correct entries in the brackets below;

(a) (A/B) lives closer to the school than (B/A)

(b) (A/B) starts from the school earlier than (B/A)

(c) (A/B) walks faster than (B/A)

(d) A and B reach home at the (same/different) time

(e) (A/B) overtakes (B/A) on the road (once/twice).

• A lives closer to school than B because from the graph it is clear that OP<OQ, Hence, the distance of school from the A’s home is less than that from B’s home.
• For A, x=0, t=0 but for B, x =0, t has some finite value. Therefore A starts his journey from school earlier than B.
• From the graph, it is clear that slope of B > slope of A. As the slope of the (x-t) graph gives the speed, this shows speed of B is greater than A.
• It is very clear from the graph that both A and B reach their respective homes at same time.
• From the graph we can say that B overtakes A only once as B’s speed is greater than that of A.

Question 3.

A woman starts from her home at 9.00 am, walks with a speed of 5 km h–1 on a straight road up to her office 2.5 km away,

stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km h–1. Choose suitable scales and plot the x-t graph of her motion.

Given:

Speed of the woman = 5 km/h

Distance between her office and home = 2.5 km

Time taken = (Distance)/ (Speed)

= (2.5/5) = 0.5 h= 30min

It is also given that she covers the same distance in the evening by an auto.

Now, speed of the auto = 25 km/h

Time taken = (Distance)/ (Speed)

= (2.5)/ (2.5) = (1/10) =0.1h = 6min

=>The woman will reach home by 5.06pm.

The suitable x-t graph represents the motion of the woman.

Question 4.

A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward,

and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion.

Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.

Given:-

Distance covered with 1 step = 1 m

Time taken = 1 s

Time taken to move first 5 m forward = 5 s

Time taken to move 3 m backward = 3 s

Net distance covered = (5 – 3) = 2 m

Net time taken to cover 2 m = 8 s

Drunkard covers 2 m in 8 s.

Drunkard covered 4 m in 16 s.

Drunkard covered 6 m in 24 s.

In the next 5 s, the drunkard will cover a distance of 5 m and a total distance of 13 m and falls into the pit.

Net time taken by the drunkard to cover 13 m = 32 + 5 = 37 s

The x-t graph of the drunkard’s motion can be shown as:

Question 5.

A jet airplane travelling at the speed of 500 km h–1 ejects its products of combustion at the speed of 1500 km h–1 relative to the jet plane.

What is the speed of the latter with respect to an observer on the ground?

Given:

Let the velocity of jet = (VJ) = 500 km/hr

Let the velocity of observer on ground (VO) = 0 km/hr

Let the velocity of ejected gas = VC.

Therefore, Relative velocity of jet w.r.t. to ground = (VJ – VO) = 500 km/hr

=> Relative velocity of ejected gas wrt jet = (VC – VJ) = -1500 km/hr

=> Relative speed of ejected gas w.r.t ground = (VC – VO) = (VC – VJ) + (VJ – VO)

= -1500 + 500 = -1000 km/hr

The (-ive) sign indicates the relative speed of ejected products is opposite to the direction of jet.

Question 6.

A car moving along a straight highway with speed of 126 km h–1 is brought to a stop within a distance of 200 m.

What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?

Given,

Initial velocity (u) of car = 126 km/hr = 126 x (1000/3600) = 35 m/s

Final velocity (v) = 0 m/s

Distance covered (s) = 200m

Using equation (v2 – u2) = 2as

a = (v2 – u2)/2s

= (0 – (35)2)/ (2 x 200)

= – (1225/400)

Using equation, v = u + at

t = (v-u)/a = (-35/-3.06) = 11.44s

It will take 11.44s for the car to stop.

Question7.

Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h–1 in the same direction, with A ahead of B.

The driver of B decides to overtake A and accelerates by 1 m s–2.

If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?

Originally, both the trains have the same velocities. So the relative velocity of B w.r.t A is 0.

• For train A:

Initial velocity, u = 72 km/h = 20 m/s

Time, t = 50 s

Acceleration, a1 = 0 (Since it is moving with a uniform velocity)

Using, second equation of motion,

Distance covered by train A s1 = ut + (1/2) a1t2

= (20 × 50) + 0 = 1000 m

• For train B:

Initial velocity, u = 72 km/h = 20 m/s

Acceleration, a = 1 m/s2

Time, t = 50 s

From second equation of motion,

s2 = ut + (1/2) at2

= (20 x 50) + (1/2) x 1 x (50)2

=2250m

Length of both the trains = (2 x 400)m =800m

Therefore, the original distance between the driver of train A and the guard of train B is (2250 –1000 - 800) = 450 m

Question 8.

On a two-lane road, car A is travelling with a speed of 36 km h–1. Two cars B and C approach car A in opposite directions with a speed of 54 km h–1 each.

At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does.

What minimum acceleration of car B is required to avoid an accident?

Given:-

Velocity of car A, vA = 36 km/h = 10 m/s

Velocity of car B, vB = 54 km/h = 15 m/s

Velocity of car C, vC = 54 km/h = 15 m/s

Relative velocity of car B with respect to car A,

vBA = vB – vA

= (15 – 10) = 5 m/s

Relative velocity of car C with respect to car A,

vCA = vC – (– vA)

= 15 + 10 = 25 m/s

At a certain instance, both cars B and C are at the same distance from car A i.e.

s = 1 km = 1000 m

Time taken (t) by car C to cover 1000 m = (1000/25) = 40s

Hence, to avoid an accident, car B must cover the same distance in a maximum of 40 s.

From second equation of motion, minimum acceleration (a) produced by car B can be obtained as:

s= ut + (1/2) at2

1000 = (5 x 40) + (1/2) x a x (40)2

a= (1600/1600) = 1ms-2

Question 9.

Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes.

A man cycling with a speed of 20 km h–1 in the direction A to B notices that a bus goes past him every 18 min in the direction of his motion,

and every 6 min in the opposite direction. What is the period T of the bus service and with what speed (assumed constant) do the buses ply on the road?

Let V be the speed of the bus running between towns A and B.

Speed of the cyclist, vc = 20 km/h

Case 1:- Relative speed of the bus moving in the direction of the cyclist

= (V – vc) = (V – 20) km/h

Since the bus goes past the cyclist every 18 min i.e., (18/60),

Therefore, the distance covered by the bus w.r.t the cyclist

= (V – 20) x (18/60) km (i)

As the bus leaves after every T minutes, the distance travelled by the bus will be

= (V x (T/60))   (ii)

As both equations (i) and (ii) are equal.

= (V – 20) x (18/60) = (VT)/ (60)   (iii)

Case 2:- Relative speed of the bus coming from town q to p w.r.t cyclist

= (V+20) kmh-1

Time taken by the bus to go past cyclist = 6 min = (6/60) h

Therefore (V +20) (6/60) = (VT)/ (60)    (iv)

From equations (iii) and (IV)

(V+20) (6/60) = (V – 20) x (18/60)

(V + 20) = 3V – 60

2V = 80

V = 40km/h

Substituting the value of V in equation (iv), we get

(40 + 20) x (6/60) = (40T)/ (60)

T = (360/40) = 9min

Question 10.

A player throws a ball upwards with an initial speed of 29.4 m s–1.

(a) What is the direction of acceleration during the upward motion of the ball?

(b) What are the velocity and acceleration of the ball at the highest point of its motion?

(c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis,

and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.

(d) To what height does the ball rise and after how long does the ball return to the player’s hands? (Take g = 9.8 m s–2 and neglect air resistance).

• Direction of ball will be always vertically downward because the ball is moving under the effect of gravity.
• At the highest point, the vertical velocity becomes 0, and the value of acceleration due to gravity is (-9.8m/s2) acting vertically downward.
• When the highest point is chosen as the location for x=0 and t=0 and vertically downward direction to be the positive direction x-axis
• and upward direction as negative direction of x-axis.

During upward motion, sign of position is negative, sign of velocity is negative and sign of acceleration is positive.

During downward motion, sign of position is positive, sign of velocity is positive and sign of acceleration is also positive.

• During upward motion;

Time taken by the ball to reach the highest point = t.

Height of the highest point from the ground = h

Given:

Initial velocity u =-29.4m/s2

Acceleration due to gravity g = 9.8m/s2

Final velocity = v

Using (v2 – u2) =2as

=> (0)2 – (29.4)2 = 2 x9.8 x h

Or h = - (29.4 x 29.4)/ (2 x 9.8)

=-44.1m   where (-) ive sign indicates that the distance is in upward direction.

Using; v =u + at

0= (-29.4) + 9.8 x t

Therefore t = (29.4)/ (9.8) = 3s.

Total time taken by the ball to return in the player’s hand

= (3s + 3s) = 6sec (because time of ascent = time of descent)

Question 11.

Read each statement below carefully and state with reasons and examples, if it is true or false;

A particle in one-dimensional motion

(a) with zero speed at an instant may have non-zero acceleration at that instant

(b) with zero speed may have non-zero velocity,

(c) with constant speed must have zero acceleration,

(d) with positive value of acceleration must be speeding up.

• True. When a body is thrown vertically upwards in the space, at the highest point speed of the body is 0.
• But the acceleration will be equal to acceleration due to gravity (g) that acts in the downward direction.
• As velocity is the speed of a body in a given direction. Therefore if speed is 0, the magnitude of velocity will also be 0.
• When a particle, is moving along a straight line with constant speed, its velocity remains constant with time. Therefore acceleration is 0.
• It will be only true when both velocity and acceleration are positive, at that instant of time taken as origin.
• Such a case happens when a particle is moving with positive acceleration or falling vertically downwards from a height.

Question 12.

A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed.

Plot the speed-time graph of its motion between t = 0 to 12 s.

Given:-

Ball is dropped from a height, s =90m

Initial velocity of the ball, u = 0

Acceleration, a = g = 9.8 m/s

Final velocity of the ball = v

From second equation of motion,

s = ut+ (1/2) at2

90=0 + (1/2) x 9.8(t2)

t = √18.38 = 4.29s

From first equation of motion, final velocity is given

v = u + at

= 0 + (9.8 × 4.29) = 42.04 m/s

Rebound velocity of the ball, ur = (9/10) v= (9/10) x 42.04 =37.84m/s

Time (t) taken by the ball to reach maximum height can be obtained by using First equation of motion as:

v = ur + at′

0 = 37.84 + (– 9.8) t′

t’ = (-37.84)/(-9.8) = 3.86s

Total time taken by the ball = (t + t’) = (4.29+3.86) =8.15s.

As the time of ascent is equal to the time of descent, the ball takes 3.86 s to strike back on the floor for the second time.

The velocity with which the ball rebounds from the floor = ((9/10) x 37.84) =34.05m/s

Total time taken by the ball for second rebound = 8.15 + 3.86 = 12.01 s

The speed-time graph of the ball is represented in the given figure as:

Question 13.

Explain clearly, with examples, the distinction between:

(a) Magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval;

(b) Magnitude of average velocity over an interval of time, and the average speed over the same interval.

[Average speed of a particle over an interval of time is defined as the total path length divided by the time interval].

Show in both (a) and (b) that the second quantity is either greater than or equal to the first.

When is the equality sign true? [For simplicity, consider one-dimensional motion only].

• Suppose a particle moves from A to B and B to C in time t as shown in the figure.

Therefore magnitude of displacement of the particle =AC

Total path covered will be = (AB+AC)

But total path length (AB +BC) is > magnitude of displacement (AC)

=> If a particle moves in a straight line, then the magnitude of displacement = total path length covered by the particle in given time.

(b)     Magnitude of average velocity = (magnitude of displacement)/ (time interval)

= (AC)/ (t) = (total path length)/ (time interval) = (AB+BC)/ (t)

(AB +BC) > (AC). If the particle moves in a straight line, then in a given time the magnitude of displacement is equal to total path transverse by a particle in a given time,

so average speed = average velocity.

Therefore in (a) or (b) the second quantity is either greater than or equal to first quantity.

Question 14.

A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h–1.

Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h–1. What is the

(a) Magnitude of average velocity, and

(b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min ?

[Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time,

and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speeds was zero!]

Given:

Time taken by the man to reach the market from home,

t1 = (2.5/5) = (1/2) h = 30min

Time taken by the man to reach home from the market,

t2= (2.5/7.5) = (1/3) h = 20min

Total time taken in the whole journey = 30 + 20 = 50 min

Average velocity = (displacement)/ (time) = (2.5)/ (1/2) =5km/h

Average speed = (distance)/ (time) = (2.5)/ (1/2) = 5km/h

Time = 50 min = (5/6) h

Net displacement = 0

Total distance = (2.5 + 2.5) = 5 km

Average velocity = (displacement)/ (time) = 0

Average speed = (distance)/ (time) = (5)/ (5/6) =6km/h

Speed of the man = 7.5 km

Distance travelled in first 30 min = 2.5 km

Distance travelled by the man (from market to home) in the next 10 min

= (7.5) x (10/60) = 1.25km

Net displacement = (2.5 – 1.25) = 1.25 km

Total distance travelled = (2.5 + 1.25) = 3.75 km

Average velocity = (1.25)/ (40/60) = (1.25 x 3)/ (2) = 1.875km/h

Average speed = (3.75)/ (40/60) = 5.625km/h

Question 15.

In Exercises 3.13 and 3.14, we have carefully distinguished between average speed and magnitude of average velocity.

No such distinction is necessary when we consider instantaneous speed and magnitude of velocity.

The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?

Instantaneous velocity is given by first derivative of distance with respect to time i.e.  v= (dx/dt)

As time interval is so small so it is assumed that particles does not change its direction of motion.

As a result there is no difference between total path length and magnitude of displacement for that small interval of time dt.

Therefore instantaneous speed is always equal to instantaneous velocity.

Question 16.

Look at the graphs (a) to (d) (Fig. 3.20) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.

• In first case, the x-t graph does not represent one-dimensional motion of the particle. Because particle cannot have two positions at the same instant of time.
• In second case, the v-t graph does not represent one-dimensional motion of the particle. This is because a particle can never have two values of velocity at the same instant of time.
• In third case, the v-t graph does not represent one particle. Because speed being a scalar quantity cannot be negative.
• In fourth case, the v-t graph does not represent one particle. This is because the total path length travelled by the particle cannot decrease with time.

Question 17.

Figure 3.21 shows the x-t plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t >0?

If not, suggest a suitable physical context for this graph.

No. As the x-t graph of a particle moving in straight line where t0 cannot be shown by the given graph.

This is because, the given particle does not follow the trajectory of path followed by the particle as t=0, x=0.

A physical situation that resembles the above graph is of a freely falling body held for some time at a height.

Question 18.

A police van moving on a highway with a speed of 30 km h–1 fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km h–1.

If the muzzle speed of the bullet is 150 m s–1, with what speed does the bullet hit the thief’s car? (Note: Obtain that speed which is relevant for damaging the thief’s car).

Given:

Speed of the police van, vp = 30 km/h = 8.33 m/s

Muzzle speed of the bullet, vb=150m/s

Speed of the thief’s car, vt = 192 km/h = 53.33 m/s

Since the bullet is fired from a moving van,

Therefore resultant speed = vbt = (vb – vt) = (150 + 8.33) = 158.33 m/s

Since both the vehicles are moving in the same direction, the velocity with which the bullet hits the thief’s car= vbt = (vb – vt)

= (158.33 – 53.33) = 105 m/s

Question 19.

Suggest a suitable physical situation for each of the following graphs (Fig 3.22):

• The given x-t graph shows that initially a body was at rest. Then, its velocity increases with time and attains an instantaneous constant value.
• The velocity then reduces to zero with an increase in time. Then, its velocity increases with time in the opposite direction and acquires a constant value.
• A similar physical situation arises when a football (initially kept at rest) is kicked and gets rebound from a rigid wall so that its speed gets reduced.
• Then, it passes from the player who has kicked it and ultimately gets stopped after sometime.
• In the given v-t graph, the sign of velocity changes and its magnitude decreases with a passage of time.
• A similar situation arises when a ball is dropped on the hard floor from a height. It strikes the floor with some velocity and upon rebound, its velocity decreases by a factor. This continues till the velocity of the ball eventually becomes zero.
• The given a-t graph reveals that initially the body is moving with a certain uniform velocity.
• Its acceleration increases for a short interval of time, which again drops to zero.
• This indicates that the body again starts moving with the same constant velocity. A similar physical situation arises when a hammer moving with a uniform velocity strikes a nail.

Question 20.

Figure 3.23 gives the x-t plot of a particle executing one-dimensional simple harmonic motion.

Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, – 1.2 s.

In simple harmonic oscillator acceleration is given by

a = (-ω2 x) where ω = angular frequency and is constant.

At t=0.3sec; It is clear from the graph that x is negative.

The slope of x-t graph at is negative, therefore velocity is also negative.

As a = (-ω2 x) therefore (a) is (+) ive.

Similarly at t=1.2 s, x is positive, v is positive and a is negative.

Similarly at t= 1.2s, x is negative, v is positive and a is negative

Question 21.

Figure 3.24 gives the x-t plot of a particle in one-dimensional motion.

Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least?

Give the sign of average velocity for each interval.

It is greatest in (interval 3) and least in (interval 2).

Positive in intervals 1 and 2 and negative in interval 3

The average speed of a particle shown in the given x-t graph will be given by the slope of the graph in a particular interval of time.

Slope is maximum in interval 3 and minimum in interval 2.

Therefore, average speed of the particle is greatest in interval 3 and least in interval 2.

As the slope is positive in interval 2and 1 therefore average velocity will also be positive.

The slope is negative in interval 3 therefore it is negative in this interval.

Question 22.

Figure 3.25 gives a speed-time graph of a particle in motion along a constant direction.

Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude?

In which interval is the average speed greatest?

Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals.

What are the accelerations at the points A, B, C and D?

Average acceleration is greatest in interval 2

Average speed is greatest in interval 3

v is positive in intervals 1, 2, and 3

a is positive in intervals 1 and 3 and negative in interval 2

a = 0 at A, B, C, D

Acceleration is given by the slope of the speed-time graph.

In the given case, it is given by the slope of the speed-time graph within the given interval of time.

Since the slope of the given speed-time graph is maximum in interval 2, average acceleration will be the greatest in this interval.

Height of the curve from the time-axis gives the average speed of the particle. It is clear that the height is the greatest in interval 3.

Hence, average speed of the particle is the greatest in interval 3.

In interval 1:

The slope of the speed-time graph is positive. Hence, acceleration is positive. Similarly, the speed of the particle is positive in this interval.

In interval 2:

The slope of the speed-time graph is negative. Hence, acceleration is negative in this interval. However, speed is positive because it is a scalar quantity.

In Interval 3:

The slope of the speed-time graph is zero. Hence, acceleration is zero in this interval.

However, here the particle acquires some uniform speed. It is positive in this interval.

Points A, B, C, and D are all parallel to the time-axis. Hence, the slope is zero at these points. Therefore, at points A, B, C, and D, acceleration of the particle is zero