Question

A body begins to move with an initial velocity of $2\,m{{s}^{-1}}$ and continues to move at a constant acceleration a. Ten seconds later a second body begins to move from the same point with an initial velocity of $12\,m{{s}^{-1}}$ in the same direction with the same acceleration. What is the maximum acceleration at which it would be possible for the second body to overtake the first?

Answer 1

Hint:To solve this problem, you have to know about Newton’s equation of motion. First, we have to formulate an equation satisfying the conditions specified in the question from which it is possible to find the maximum acceleration.

Formula used:
Newton’s equations of motion are used. Let a be the acceleration of the particle, v be the final velocity and u be the initial velocity of the particle. Let t be the time taken by the particle and s be the distance travelled by the particle, then Newton’s equation of motion is:
1. $v=u+at$
2.$s=ut+\dfrac{1}{2}a{{t}^{2}}$
3.${{v}^{2}}={{u}^{2}}+2as$

Complete step by step solution:
First body travels with an initial velocity $2m{{s}^{-1}}$ and then it moves with a constant acceleration a. The distance covered by the first body in ten seconds according to Newton’s second equation of motion is:
$s=2\times 10+\dfrac{1}{2}a\times 100=20+50a$.................equation (1)

Let the two bodies meet at t seconds after the second starts. Then distance covered by the second particle in t seconds is:
$s'=12t+\dfrac{1}{2}a{{t}^{2}}$
The distance covered by the first body from the starting point is:
$s=(20+50a)+(2+10a)t+\dfrac{1}{2}a{{t}^{2}} \\
\Rightarrow s=20+50a+10at+2t+\dfrac{1}{2}a{{t}^{2}}$..............equation (2)
We have to find the maximum acceleration at which the second body overtakes the first. For overtaking the distance covered by both bodies should be the same.

Therefore, equating (1) and (2) we get:
$20+50a+10at+2t+\dfrac{1}{2}a{{t}^{2}}=12t+\dfrac{1}{2}a{{t}^{2}} \\ $
We can simplify it as:
$20+50a+10at-10t=0$
Time, $t=\dfrac{5a+2}{(1-a)} \\ $
We know that time cannot be negative.
That is, $1-a\ge 0$

Hence, the value of acceleration is, $a=1\,m{{s}^{-2}}$.

Note: It is given in the question that both particles travel in the same direction with same acceleration. Time cannot be negative. While making equations and simplifying it you have to be very careful.