Search any question & find its solution
Question:
Answered & Verified by Expert
A cyclist is riding with a speed of $36 \mathrm{~km} / \mathrm{h}$. As he approaches a circular turn on the road of radius $50 \mathrm{~m}$, he applies brakes and reduces his speed at the constant rate of $0.5 \mathrm{~m} / \mathrm{s}$ every second. The magnitude and direction of the net acceleration of the cyclist on the circular turn respectively are
Options:
Solution:
2104 Upvotes
Verified Answer
The correct answer is:
$\frac{\sqrt{17}}{2} \mathrm{~m} / \mathrm{s}^2, \tan ^{-1}(4)$
We have
$$
\begin{aligned}
& \mathrm{a}_{\mathrm{r}}=\frac{\mathrm{V}^2}{\mathrm{R}}=\frac{10^2}{50}=2 \mathrm{~m} / \mathrm{s}^2 \\
& \mathrm{a}_{\mathrm{t}}=0.5 \mathrm{~m} / \mathrm{s}^2 \\
& \text { So, } \mathrm{a}_{\text {net }}=\sqrt{2^2+0.5^2}=\sqrt{\frac{17}{4}}=\frac{\sqrt{17}}{2} \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
Now, $\tan \theta=\frac{\mathrm{a}_{\mathrm{r}}}{\mathrm{a}_{\mathrm{t}}}=\frac{2}{0.5}=4 \Rightarrow \theta=\tan ^{-1}(4)$
$$
\begin{aligned}
& \mathrm{a}_{\mathrm{r}}=\frac{\mathrm{V}^2}{\mathrm{R}}=\frac{10^2}{50}=2 \mathrm{~m} / \mathrm{s}^2 \\
& \mathrm{a}_{\mathrm{t}}=0.5 \mathrm{~m} / \mathrm{s}^2 \\
& \text { So, } \mathrm{a}_{\text {net }}=\sqrt{2^2+0.5^2}=\sqrt{\frac{17}{4}}=\frac{\sqrt{17}}{2} \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
Now, $\tan \theta=\frac{\mathrm{a}_{\mathrm{r}}}{\mathrm{a}_{\mathrm{t}}}=\frac{2}{0.5}=4 \Rightarrow \theta=\tan ^{-1}(4)$
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.