Search any question & find its solution
Question:
Answered & Verified by Expert
A particle moving in a straight line starts from rest and the acceleration at any time $\mathrm{t}$ is $a-k t^2{ }^2$ where a and $k$ are positive constants. The maximum velocity attained by the particle is
Options:
Solution:
1691 Upvotes
Verified Answer
The correct answer is:
$\frac{2}{3} \sqrt{a^3 / k}$
$\frac{d v}{d t}=a-k t^2=(\sqrt{a}-\sqrt{k t})(\sqrt{a}+\sqrt{k t})[\because a, k>0]$
$v$ will be $\max$ at $t=\sqrt{\frac{a}{k}}$
$$
\because v=a t-\frac{k}{3} t^3+C
$$
at $\mathrm{t}=0, \mathrm{v}=0 \Rightarrow \mathrm{C}=0$
$\begin{aligned} \therefore v &=a t-\frac{k}{3} t^3 \\ \therefore v_{\max } &=a \sqrt{\frac{a}{k}}-\frac{k}{3} \frac{a}{k} \sqrt{\frac{a}{k}} \\ &=\frac{2 a}{3} \sqrt{\frac{a}{\mathrm{k}}} \end{aligned}$
$v$ will be $\max$ at $t=\sqrt{\frac{a}{k}}$
$$
\because v=a t-\frac{k}{3} t^3+C
$$
at $\mathrm{t}=0, \mathrm{v}=0 \Rightarrow \mathrm{C}=0$
$\begin{aligned} \therefore v &=a t-\frac{k}{3} t^3 \\ \therefore v_{\max } &=a \sqrt{\frac{a}{k}}-\frac{k}{3} \frac{a}{k} \sqrt{\frac{a}{k}} \\ &=\frac{2 a}{3} \sqrt{\frac{a}{\mathrm{k}}} \end{aligned}$
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.