WBJEE · Maths · Application of Derivatives
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
- A \(\frac{2}{3} \sqrt{a^3 / k}\)
- B \(\frac{1}{3} \sqrt{a^3 / k}\)
- C \(\sqrt{a^3 / k}\)
- D \(2 \sqrt{a^3 / k}\)
Answer & Solution
Correct Answer
(A) \(\frac{2}{3} \sqrt{a^3 / k}\)
Step-by-step Solution
Detailed explanation
\(\frac{d v}{d t}=a-k t^2=(\sqrt{a}-\sqrt{k t})(\sqrt{a}+\sqrt{k t})[\because a, k>0]\) \(v\) will be \(\max\) at \(t=\sqrt{\frac{a}{k}}\) \[ \because v=a t-\frac{k}{3} t^3+C \] at \(\mathrm{t}=0, \mathrm{v}=0 \Rightarrow \mathrm{C}=0\)…
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