Question

A particle moves in a straight line and passes through O, a fixed point on the line, with a velocity of $6m{{s}^{-1}}$. The particle moves with constant retardation of $2m{{s}^{-2}}$ for 4 seconds and thereafter moves with constant velocity. How long after leaving O does the particle return to O?
A. $8s$
B. $2s$
C. $4s$
D. $6s$

Answer 1

Hint: When a deacceleration is acting on a moving body, it eventually stops the body. If the deacceleration is still present after the body is halted, the deacceleration will act as an acceleration in the opposite direction of the initial motion. After the duration of acceleration is over, the body will start to move with a constant velocity.

Complete step by step answer:

A particle denoted by the square in the figure has a velocity of $6m{{s}^{-1}}$ when it passes the point O. A constant deacceleration is acting on the particle for a duration of 4 seconds. So the deacceleration acting on the body will eventually stop the body such that its final velocity will be zero at point L. So, the time taken by the deacceleration to halt the object is given by the formula,
$v=u-at$
Here v=0, so we can write,
$t=\dfrac{u}{a}=\dfrac{6m{{s}^{-1}}}{2m{{s}^{-2}}}$
$\therefore {{t}_{1}}=3\text{ seconds}$… equation (1)
So, the time taken for the particle to stop is 3 seconds.
The distance travelled in these 3 seconds is given by the formula, $s=ut-\dfrac{1}{2}a{{t}^{2}}$, Substituting the values of u, t and a in the equation we get,
${{s}_{1}}=(6\times 3)-\dfrac{1}{2}\left( 2\times 9 \right)$
$\therefore {{s}_{1}}=9m$
It is the given, in the problem that the deacceleration acts on the body for a period of 4 seconds. So, 3 seconds is spent to stop the body, and the other one second is used to accelerate the body in the opposite direction of its initial motion. So, the velocity gained in a time period of 1 second can be found out by using the formula, $v=u+at$, here u- which is the initial velocity is zero. So, we can write,
$v=(2m{{s}^{-2}})\times 1s$
$\therefore v=2m{{s}^{-1}}$ …equation (2)
The distance travelled in the direction of the deacceleration is given by the formula,
$s=\dfrac{1}{2}a{{t}^{2}}$
We know that t=1 second and u=0. So, substituting the value in the above equation, we get,
${{s}_{2}}=\dfrac{1}{2}\left( 2m{{s}^{-2}} \right)\times {{\left( 1s \right)}^{2}}$
$\therefore {{s}_{2}}=1m$ … equation (3)
So, the particle travels 1 metre with acceleration $2m{{s}^{-2}}$and travels rest of the distance which is ${{s}_{3}}=(9-1)m=8m$,(from equation (3)), with a constant velocity of $v=2m{{s}^{-1}}$. (From equation (2))
So, the time taken to travel the rest of the distance with a constant velocity of $v=2m{{s}^{-1}}$ is given by,
$\text{Time}=\dfrac{\text{Distance}}{\text{Velocity}}$
$\text{Time(}{{\text{t}}_{3}})=\dfrac{{{s}_{3}}}{v}=\dfrac{8m}{2m{{s}^{-1}}}$
$\therefore {{t}_{3}}=4s$ …. equation (4)
So, the total time taken to reach the point again is, $T={{t}_{1}}+{{t}_{2}}+{{t}_{3}}$
$\Rightarrow T=3s+1s+4s$
$\therefore T=8\text{ seconds}$
So, the total time taken is 8 seconds.
So, the answer to the question is option (A).

Note: Newton’s law of motion is defined for motion is only valid for the motion of a body having constant acceleration, that is the acceleration of the body should not change with time. In the given problem, the body does not have a constant acceleration, and it varies linearly with time.
The velocity of a body is defined as the rate of change of displacement of the body.
The slope of the displacement time graph gives the velocity of the body, while the slope of the velocity-time graph gives the acceleration of the body.
The area under the velocity-time graph gives the displacement of the body in that period of time. While the area under the acceleration-time graph gives the velocity of the body in that period of time.