Question

A particle starts from origin with a velocity of 10 m/s and moves with a constant acceleration till the velocity increases to 50 m/s. At that instant, the acceleration is suddenly reversed. What will be the velocity of the particle, when it returns to the starting point?
A) zero
B) 10 m/s
C) 50 m/s
D) 70 m/s

Answer 1

Hint: The important part to understand while solving this problem is the fact that the distance travelled with the forward motion and the distance travelled with the reverse motion must be calculated in order to compute the velocity in the backward direction.

Complete step by step answer:
When a particle starts from origin from initial velocity of 10 m/s to a final velocity of 50 m/s under uniform acceleration, the distance travelled is given by the equation –
$\Rightarrow {v^2} - {u^2} = 2a{s_1}$
To calculate the distance travelled in the forward motion, ${s_1}$ –
$\Rightarrow {50^2} - {10^2} = 2a{s_1}$
$ \Rightarrow 2a{s_1} = 2500 - 100$
$ \Rightarrow {s_1} = \dfrac{{2400}}{{2a}}$
At the point when it reaches a velocity of 50 m/s, the acceleration is reversed. This results in a backward motion with the initial velocity of 50 m/s and the final velocity being equal to 0 m/s.
The particle then travels a certain distance in the backward direction in negative value of acceleration. This distance is given by –
$\Rightarrow {v^2} - {u^2} = - 2a{s_2}$
$ \Rightarrow {0^2} - {50^2} = - 2a{s_2}$
$ \Rightarrow {s_2} = \dfrac{{2500}}{{2a}}$
The net displacement of the particle gives us the value of how far the particle lies from the origin and is equal to the sum distance travelled in the forward direction and the backward direction.
The net displacement,
$\Rightarrow d = {s_1} + {s_2}$
Substituting the values of ${s_1}$ and ${s_2}$ , we get –
$\Rightarrow d = \dfrac{{2400}}{{2a}} + \dfrac{{2500}}{{2a}}$
$ \Rightarrow d = \dfrac{{2400 + 2500}}{{2a}}$
$ \Rightarrow d = \dfrac{{4900}}{{2a}}$
Given that the particle starts from zero and reaches a point whose displacement d is given by the above expression, the final velocity achieved while returning to its starting point is given by –
$\Rightarrow {v^2} - {u^2} = 2ad$
Since the particle starts from rest, $u = 0$
$ \Rightarrow {v^2} = 2ad$
Substituting the value of d, we get –
$ \Rightarrow {v^2} = 2a \times \dfrac{{4900}}{{2a}}$
$ \Rightarrow {v^2} = 4900$
$\therefore v = 70m{s^{ - 1}}$

Hence, the correct option is Option D.

Note: In this problem, we can clearly understand the difference between the quantities distance and displacement. We can see here that even though the total distance travelled by the particle is equal to the sum of ${s_1} + {s_2}$ , we can infer from above that, the net displacement of the object is not greater than the distance ${s_1}$.