$\begin{array}{1 1}1.6m/s^2\\2.6m/s^2\\3.6m/s^2\\4.6m/s^2\end{array} $

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Answer : $1.6m/s^2$

The period is $T=\large\frac{1}{4}$ min=15s

A rider's speed is $v=\large\frac{2\pi r}{T}$

$\Rightarrow \large\frac{2\pi (9.0m)}{15s}$

$\Rightarrow 3.77m/s$

Consequently,the centripetal acceleration is

$a=\large\frac{v^2}{r}$

$\Rightarrow \large\frac{(3.77m/s)^2}{9.0m)}$

$\Rightarrow 1.6m/s^2$

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...