Search any question & find its solution
Question:
Answered & Verified by Expert
A cyclist starts from centre $O$ of a circular park of radius 1 $\mathrm{km}$ and moves along the path $O P R Q O$ as shown in figure. If he maintains constant speed of $10 \mathrm{~ms}^{-1}$, what is his acceleration at point $R$ in magnitude and direction?
Solution:
2875 Upvotes
Verified Answer
According to given figure, the cyclist covers the path $O P R Q O$. When an object performing circular motion, acceleration is called centripetal acceleration and is always directed towards the centre.
As given that
$$
\begin{aligned}
&R=1 \mathrm{~km}=1000 \mathrm{~m} \\
&v=10 \mathrm{~m} / \mathrm{s} .
\end{aligned}
$$
So, acceleration at $R$ is $a=\frac{v^2}{r}$
$$
a=\frac{(10)^2}{1 \mathrm{~km}}=\frac{100}{10^3}=0.1 \mathrm{~m} / \mathrm{s}^2 \text { along } R O
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.