Question

A scooterist is approaching a circular turn of radius $80m$. He reduced his speed from $27km/h$ at constant rate of $0.5m/s^2$. His vector acceleration on the circular turn is:

• A. $0.86m/s^2$, ${54}^{\circ }$
• B. $0.68m/s^2$, ${45}^{\circ }$
• C. $1.0m/s^2$, ${45}^{\circ }$
• D. $0.5m/s^2$, ${45}^{\circ }$

Answer 1

The correct option is A $0.86m/s^2$, ${54}^{\circ }$

Centripetal acceleration
$a_r=\frac{v^2}{r}={\left(\frac{27\times 1000}{3600}\right)}^2\times \frac{1}{80}$$=\frac{7.5\times 7.5}{80}=0.703m/s^2$

Net acceleration, $a=\sqrt{a_r^2+a_t^2}$, where $a_t$ = tangential acceleration = $0.5m/s^2$

So, $a=\sqrt{{0.703}^2+{0.5}^2}=0.86m/s^2$ and

$\theta ={\tan }^{-1}\frac{a_r}{a_t}=ta{n}^{-1}\frac{0.703}{0.5}={54}^{\circ }$