Question

A car is travelling along a circular curve that has a radius of $50\text{m}$ with a constant tangential acceleration. If the speed of the car changes from $36\text{km/hr}$ to $45\text{km/hr}$ in $5\text{seconds}$, the angular acceleration of the car is.(No slipping between road and car tyre)

• A. $0.05{\text{rad/s}}^2$
• B. $0.01{\text{rad/s}}^2$
• C. $0.02{\text{rad/s}}^2$
• D. $0.04{\text{rad/s}}^2$

Answer 1

The correct option is B $0.01{\text{rad/s}}^2$

We know that,
$v=r\omega$
$\Rightarrow a=r\alpha$
Since the car travels with a constant tangential acceleration, we have
$v=u+at$
$\Rightarrow a=\frac{v-u}{t}$
$v=45\text{km/hr}=45\times \frac{5}{18}=12.5\text{m/s}$
$u=36\text{km/hr}=36\times \frac{5}{18}=10\text{m/s}$
$\therefore a=\frac{v-u}{t}=\frac{12.5-10}{5}=0.5{\text{m/s}}^2$
From $a=r\alpha$, angular acceleration
$\alpha =\frac{a}{r}=\frac{0.5}{50}=0.01{\text{rad/s}}^2$