Question

The velocity of a car travelling on a straight road is $36km{h}^{-1}$ at an instant of time. Now travelling with uniform acceleration for 10 s, the velocity becomes exactly double. If the wheel radius of the car is 25 cm, then which of the following numbers is the closest to the number of revolutions that the wheel makes during this 10 s?

• A. $84$
• B. $95$
• C. $126$
• D. $135$

Answer 1

Answer: (B)

$95$

Radius of wheel $r=25$ cm $=0.25$ m

Initial velocity of car $u=36km/h=36\times \frac{5}{18}=10m/s$

Thus initial angular speed of wheel $w_i=\frac{u}{r}=\frac{10}{0.25}=40$ $rad/s$

Final velocity of car $v=2u=20m/s$

Final angular speed of wheel $w_f=\frac{v}{r}=\frac{20}{0.25}=80$ $rad/s$

Using $w_f=w_i=\alpha t$ where $t=10$ s

$\therefore$ $80=40+\alpha \left(10\right)$ $⟹\alpha =4$ $rad/s^2$

Using $\theta =w_{i}t+\frac{1}{2}\alpha t^2$

We get $\theta =\left(40\right)\left(10\right)+\frac{1}{2}\left(4\right)(10{)}^2=600$ $rad$

Thus number of revolutions $n=\frac{\theta }{2\pi }=\frac{600}{2\pi }=95.54$ revoltions

Thus about $95$ revolution were made by the wheel.