Updated By SaraNextGen

On March 11, 2024, 11:35 AM

Question 3.22:

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?

(Fig: 3.25)

Answer:

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.

Question 3.23:

A three-wheeler starts from rest, accelerates uniformly with 1 m s^–2 on a straight road for 10 s, and then moves with uniform velocity. Plot the distance covered by the vehicle during the n^th second (n = 1,2,3….) versus n. What do you expect this plot to be during accelerated motion: a straight line or a parabola?

Answer:

Straight line

Distance covered by a body in n^th second is given by the relation

Where,

u = Initial velocity

a = Acceleration

n = Time = 1, 2, 3, ….. ,n

In the given case,

u = 0 and a = 1 m/s²

This relation shows that:

Dn ∝ n … (iii)

Now, substituting different values of n in equation (iii), we get the following table:

The plot between n and Dn will be a straight line as shown:

Since the given three-wheeler acquires uniform velocity after 10 s, the line will be parallel to the time-axis after n = 10 s.

Question 3.24:

A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 m/s. How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of 5 m/s and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands?

Answer:

Initial velocity of the ball, u = 49 m/s

Acceleration, a = – g = – 9.8 m/s²

Case I:

When the lift was stationary, the boy throws the ball.

Taking upward motion of the ball,

Final velocity, v of the ball becomes zero at the highest point.

From first equation of motion, time of ascent (t) is given as:

But, the time of ascent is equal to the time of descent.

Hence, the total time taken by the ball to return to the boy’s hand = 5 + 5 = 10 s.

Case II:

The lift was moving up with a uniform velocity of 5 m/s. In this case, the relative velocity of the ball with respect to the boy remains the same i.e., 49 m/s. Therefore, in this case also, the ball will return back to the boy’s hand after 10 s.

Question 3.25:

On a long horizontally moving belt (Fig. 3.26), a child runs to and fro with a speed 9 km h^–1 (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h^–1. For an observer on a stationary platform outside, what is the

(a) speed of the child running in the direction of motion of the belt ?.

(b) speed of the child running opposite to the direction of motion of the belt ?

(c) time taken by the child in (a) and (b) ?

Which of the Answers alter if motion is viewed by one of the parents?

(Fig: 3.26)

Answer:

3.25

(a) Speed of the belt, v_B = 4 km/h

Speed of the boy, v_b = 9 km/h

Since the boy is running in the same direction of the motion of the belt, his speed (as observed by the stationary observer) can be obtained as:

v_bB = v_b + v_B = 9 + 4 = 13 km/h

(b) Since the boy is running in the direction opposite to the direction of the motion of the belt, his speed (as observed by the stationary observer) can be obtained as:

v_bB = v_b + (– v_B) = 9 – 4 = 5 km/h

(c) Distance between the child’s parents = 50 m

As both parents are standing on the moving belt, the speed of the child in either direction as observed by the parents will remain the same i.e., 9 km/h = 2.5 m/s.

Hence, the time taken by the child to move towards one of his parents is  .

(d) If the motion is viewed by any one of the parents, Answers obtained in (a) and (b) get altered. This is because the child and his parents are standing on the same belt and hence, are equally affected by the motion of the belt. Therefore, for both parents (irrespective of the direction of motion) the speed of the child remains the same i.e., 9 km/h.

For this reason, it can be concluded that the time taken by the child to reach any one of his parents remains unaltered.

Question 3.26:

Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 m/s and 30 m/s. Verify that the graph shown in Fig. 3.27 correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take g = 10 m/s². Give the equations for the linear and curved parts of the plot.

Answer:

For first stone:

Initial velocity, u_I = 15 m/s

Acceleration, a = –g = – 10 m/s²

Using the relation,

When this stone hits the ground, x₁ = 0

∴– 5t² + 15t + 200 = 0

t² – 3t – 40 = 0

t² – 8t + 5t – 40 = 0

t (t – 8) + 5 (t – 8) = 0

t = 8 s or t = – 5 s

Since the stone was projected at time t = 0, the negative sign before time is meaningless.

∴t = 8 s

For second stone:

Initial velocity, u_II = 30 m/s

Acceleration, a = –g = – 10 m/s²

Using the relation,

At the moment when this stone hits the ground; x₂ = 0

– 5t² + 30 t + 200 = 0

t² – 6t – 40 = 0

t² – 10t + 4t + 40 = 0

t (t – 10) + 4 (t – 10) = 0

t (t – 10) (t + 4) = 0

t = 10 s or t = – 4 s

Here again, the negative sign is meaningless.

∴t = 10 s

Subtracting equations (i) and (ii), we get

Equation (iii) represents the linear path of both stones. Due to this linear relation between (x₂ – x₁) and t, the path remains a straight line till 8 s.

Maximum separation between the two stones is at t = 8 s.

(x₂ – x₁)_max = 15× 8 = 120 m

This is in accordance with the given graph.

After 8 s, only second stone is in motion whose variation with time is given by the quadratic equation:

x₂ – x₁ = 200 + 30t – 5t²

Hence, the equation of linear and curved path is given by

x₂ – x₁ = 15t (Linear path)

x₂ ­– x₁ = 200 + 30t – 5t² (Curved path)