Question

A person moves along a circular thread with speed $10m{s}^{-1}$. He covers one cycle in 10 minutes.
The person is now given a challenge. He is required to complete a distance equal to the circumference in a straight line in 5 minutes. What should be the magnitude of constant acceleration that the person should maintain to complete his challenge? Assume he is starting from rest.

• A. $\frac{1}{15}$ $m{s}^{-2}$
• B. $\frac{2}{15}$ $m{s}^{-2}$
• C. $\frac{3}{15}$ $m{s}^{-2}$
• D. $\frac{4}{15}$ $m{s}^{-2}$

Answer 1

The correct option is B $\frac{2}{15}$ $m{s}^{-2}$

Given:
First case:
Speed, $v=10m{s}^{-1}$
Time taken for one cycle, $t=10min=10\times 60s=600s$
Second case,
Time taken for covering twice the distance is, $t^{\prime }=5min=5\times 60s=300s$
Initial speed, $u^{\prime }=0$

First case:
Let the radius of the circle be $R$.
Using definition of speed,
$v=\frac{2\pi R}{t}$
$R=\frac{vt}{2\pi }$

In the second case, let the acceleration be $a^{\prime }$.
Distance covered is the same. Hence, distance is:
$s=2\pi R$
Using second equation of motion,
$s=u{t}^{\prime }+\frac{1}{2}{a}^{\prime }{t}^{\prime 2}$
$2\pi R=0+\frac{1}{2}{a}^{\prime }{t}^{\prime 2}$
$a^{\prime }=\frac{4\pi R}{t^{\prime 2}}$
$a^{\prime }=\frac{2vt}{t^{\prime 2}}$
$a^{\prime }=\frac{2\times 10\times 600}{{300}^2}=\frac{2}{15}m/s^2$