Question

The rear wheels of a car are turning at an angular speed of $60$ rad/s. The brakes are applied for $5$s, causing a uniform angular retardation of $8rad/s^{-2}$. The number of revolutions turned by the rear wheels during the braking period is about:

• A. $48$
• B. $96$
• C. $32$
• D. $12$

Answer 1

The correct option is C $32$

Given : $w_i=60$ rad/s $t=5$ s $\alpha =-8$ $rad/s^2$

Using $w_f=w_i+\alpha t$

$\therefore$ $w_f=60+(-8)\times 5$

$⟹w_f=20$ rad/s

Using $w_f^2-w_i^2=2\alpha \theta$

$\therefore$ $(20{)}^2-(60{)}^2=2\times (-8)\theta$

$⟹\theta =200$ rad

Thus number of revolutions: $N=\frac{\theta }{2\pi }\approx 32$ revolutions