91Ó°ÊÓ

The angle through which the object rotates is \(\theta \).

The time taken is\(t\).

The time at which the angle is calculated is\(\frac{1}{2}t\).

In this problem, the angular velocity of the object is not constant; so the possibility of having half the distance of the rider covered in half the time is not true in this condition.

The relation from the kinematic equation is given by:

\(\theta = \omega t + \frac{1}{2}\alpha {t^2}\)

Here, \(\alpha \)is the angular acceleration and \(\omega \) is the angular velocity.

To determine the angle of the object, the above relation will be used.

\(\begin{aligned}{l}\theta ' &= \omega t + \frac{1}{2}\alpha {\left( {\frac{1}{2}t} \right)^2}\\\theta ' &= \omega t + \frac{1}{2}\alpha \left( {\frac{{{t^2}}}{4}} \right)\\\theta ' &= \frac{\theta }{4}\end{aligned}\)

Thus, option (b) is the correct answer.