Motion in Straight Line

NCERT Solution

Part 3

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.

Answer: Given, u = 0, s = 90, g = -9.8 m/s²

Time t₁ to reach ground can be calculated as follows:

`s=ut+1/2at^2`

Or, `90=0-4.9t^2`

Or, `t^2=(90)/(4.9)`

Or, `t=sqrt((900)/(49))=(30)/7=4.29` s

Velocity when the ball reaches the ground can be calculated as follows:

v² = u² + 2as

= 2 × 9.8 × 90 = 19.6 × 90

Or, v = 14 × 3 = 42 m/s

Rebound velocity u_r = 42 × 90% = 37.8 m/s

Now time (t₂) taken to reach maximum height with this rebound velocity can be calculated as follows:

v = u_r + at

Or, 0 = 37.8 – 9.8t

Or, 9.8t = 37.8

Or, `t=(37.8)/(9.8)=3.86` s

Total time after one rebound = 4.29 + 3.86 = 8.15 s

We know that time of ascent = time of descent

So, total time after ball reaches ground second time = 8.15 + 3.86 = 12.01 s

Rebound velocity on second bounce = 37.8 × 90% = 34.02 m/s

Following speed-time graph can be calculated with this data:

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.

Answer: Total distance moved by an object with change in time is called path length. The change in position of an object is called displacement

Let us choose an x-axis to understand this. In the given figure, O represents the origin. Let us assume that a car starts from O and moves to P. After that, the car moves from P to Q.

The distance covered by the car = OP + PQ

360 m + 120 m = 480 m

Displacement when the car moves from O to P

= 360 m – 0 m = 360 m

Displacement, when the car moves from P to Q

= 240 m – 360 m = -120 m

This shows magnitude of displacement may or may not be equal to the path length. Displacement can never be greater than path length.

(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).

Answer: The change in position or displacement (Δx) divided by the time intervals (Δt) is called average velocity.

v = `(x_2-x_1)/(t_2-t_1)=(Δx)/(Δt)`

where, x₂ and x₁ are positions of object respectively at time t₂ and t₁.

Average Speed: The total path length travelled divided by total time interval gives the average speed.

Average Speed = Total Path Length ÷ Total time interval

Let us take example from previous question to understand the difference.

Path length = 480 km

Displacement = 120 km

Let us assume that the car took 6 hours to travel from O to P and back to Q

Average speed = 480 ÷ 6 = 80 km/h

Average velocity = 120 ÷ 6 = 20 km/h

In this case, Average speed > Average velocity

When car moves from O to P and takes 6 hours

Average speed = 360 ÷ 6 = 60 km/h

Average velocity = 360 ÷ 6 = 60 km/h

In this case, Average speed = Average velocity

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

Answer: Since the man comes back to his home, displacement = 0, hence velocity = 0

(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?

Answer: (i) time = distance ÷ speed = 2.5 ÷ 5 = 0.5 h = 30 m so, average speed = 5 km/h

(ii) while walking back, time taken = 2.5 ÷ 7.5 = 1/3 h = 20 min

displacement in 50 m = 2.5 + 2.5 = 5 km

Speed = 5 ÷ `(50)/(60)` = 6 km/h

At the speed of 7.5 km/h he will cover 1.25 km in 10 min

So, displacement in 40 m = 2.5 + 1.25 = 3.75 km

Speed = 3.75 ÷ `(40)/(60)` = 5.625 km/h

Question 15: In previous two questions, we have carefullly 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?

Answer: Instantaneous velocity pertains to a very small duration of time. In this small time period, displacement is equal to path length. Hence, instantaneous speed is always equal to instantaneous velocity.

Question 16: Look at the graphs (a) to (d) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.

Answer:

In this case, the particle shows two different positions at the same instance of time which is not possible in one-dimensional motion.
This is v-t graph in which particle has different velocities at the same instance of time which is not possible in one-dimensional motion.
This is speed-t graph. As speed cannot be negative so this is not possible
This is total path length Vs time graph. Path length can never decrease as shown in graph. So this motion is not possible in one dimension.