Motion in Straight Line

NCERT Solution

Additional Exercise

Question 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: Distance traveled in n second is given by following equation:

`s_n=u+1/2a(2n-1)`

Given, u = 0 and a = 1 m s^-2

So, distance covered in n^th second

`s_n=0+(2n-1)/2=(2n-1)/2`

Substituting different values of n we get following values for distance:

n	1	2	3	4	5	6	7	8	9	10
s_n	0.5	1.5	2.5	3.5	4.5	5.5	6.5	7.5	8.5	9.5

This data gives following graph:

Question 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^-1. 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^-1 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: Given, u = 49 m/s, v = 0 and a = -9.8 m/s²

Time can be calculated as follows:

v = u + at

Or, 0 = 49 - 9.8t

Or, 9.8t = 49

Or, t = 49 ÷ 9.8 = 5 s

Since, time for ascent = time for descent

So, total time = 5 + 5 = 10 s

Second Case: When the lift starts moving, relative velocity of ball wrt boy remains the same, i.e. 49 m/s. So, in this case also the ball will take 10 s to reach his hand.

Question 25: On a long horizontally moving belt, 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?

Answer: 9 + 4 = 13 km/h

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

Answer: 9 – 4 = 5 km/h

Answer: Time = distance ÷ speed

`(50)/(1000)xx1/9xx60xx60=20` s

Time will be 20 s in both cases

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

Answer: Both mother and father will be able to see the relative velocity of the child wrt to belt and hence answers for (a) and (b) will change.

Question 26: Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 m s^-1 and 30 m s^-1. Verify that the graph shown in given figure 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 s^-2. Give the equations for the linear and curved parts of the plot.

Answer: Given, s₀ = 200 m, u₁ = 15 m/s, u₂ = 30 m/s a = -10 m/s^-2

For first stone:

`s_1=s_0+ut+1/2at^2`

`=200+15t-5t^2`

`=-5t^2+15t+200`

When the stone reaches ground s₁ becomes zero

So, `-5t^2+15t+200=0`

Or, -t² + 3t + 40 = 0

Or, t² - 3t – 40 = 0

Or, t² -8t + 5t – 40 = 0

Or, t(t – 8) + 5(t – 8) = 0

Or, t = -5 and t = 8

Discarding the negative value, we have t = 8 s

Similarly, t can be calculated for second stone as follows:

`s_2=-5t^2+30t+200`

Or, 5t² - 30t – 200 = 0

Or, t² - 6t – 40 = 0

Or, t² - 10t + 4t – 40 = 0

Or, t(t – 10) + 4(t – 10) = 0

So, t = 10 s

There is linear relationship between s₂ and s₁ so graph is linear up to 8 second (the time when 1st stone stops moving).

After that, equation for s₂ remains a quadratic equation and hence we get a curve for graph. So, the given graph is correct depiction.

Question 27: The speed-time graph of a particle moving along a fixed direction is shown in following figure. Obtain the distance traversed by the particle between (a) t = 0 s to 10 s, (b) t = 2 s to 6 s.

What is the average speed of the particle over the intervals in (a) and (b)

Answer: Acceleration can be calculated as follows:

Given, u =0, v = 12 m/s, t = 5 s

v = u + at

Or, 12 = 0 + 5a

Or, a = 12 ÷ 5 = 2.4 m/s²

Distance in 5 s can be calculated as follows:

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

`=0+1/2xx2.4xx5^2`

= 1.2 × 25 = 30 m

Distance covered in next 5 s (5 to 10 s) can be calculated as follows:

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

= 12 × 5 - 1.2 × 25

= 60 – 30 = 30 m

Total distance in 10 s = 30 + 30 = 60 m

Average speed = 60 ÷ 10 = 6 m/s

Distance can also be calculated by finding area of the triangle:

Area of triangle = `1/2` × base × height

= `1/2` × 12 × 10 = 60 m

Distance covered in 2 s can be calculated as follows:

S = 1.2 × 2² = 4.8 m

So, distance covered between 2 s to 5 s = 30 – 4.8 = 25.2 m

Now, distance covered in 6th second

S = 12 × 1 – 1.2 × 1 = 10.8 m

So, distance covered from 2s to 6 s = 25.2 + 10.8 = 36 m

Average speed = 36 ÷ 4 = 9 m/s

Question 28: The velocity-time graph of a particle in one-dimensional motion is shown in following figure.

Which of the following formulae are correct for describing the motion of the particle over the time interval t₁ to t₂.

x(t₂) = x(t₁) + v(t₁) (t₂ - t₁) + (½) a(t₂ - t₁)²
v(t₂) = v(t₁) + a (t₂ - t₁)
v_average = (x(t₂) – x(t₁))/( t₂- t₁)
a_average = (v(t₂) – v(t₁))/( t₂ - t₁)
x(t₂) = x(t₁) + v_average (t₂ - t₁) + (½) a_average (t₂ - t₁)²
x(t₂) – x(t₁) = area under the v-t curve bound by the t-axis and the dotted line shown.

Answer: Equations (a), (b) and (e) represent uniform acceleration. But graph shows non-uniform acceleration. Hence, equations (c), (d) and (f) are correct