Question

A particle is moving in a straight line such that its distance at any time t is given by S = $\frac{t^4}{4}-2t^3+4t^2-7.$Find when its velocity is maximum and acceleration minimum.

Answer 1

$$\begin{gathered} \mathrm{Given}:\hspace{0.17em}s=\frac{t^4}{4}-2t^3+4t^2-7 \\ \Rightarrow v=\frac{ds}{dt}=t^3-6t^2+8t \\ \Rightarrow a=\frac{dv}{dt}=3t^2-12t+8 \\ \mathrm{For}\mathrm{maximum}\mathrm{or}\mathrm{minimum}\mathrm{values}\mathrm{of}v,\mathrm{we}\mathrm{must}\mathrm{have} \\ \frac{dv}{dt}=0 \\ \Rightarrow 3t^2-12t+8=0 \\ \mathrm{On}\mathrm{solving}\mathrm{the}\mathrm{equation},\mathrm{we}\mathrm{get} \\ t=2\pm \frac{2}{\sqrt{3}} \\ \mathrm{Now}, \\ \frac{d^{2}v}{d{t}^2}=6t-12 \\ \mathrm{At}t=2-\frac{2}{\sqrt{3}}: \\ \frac{d^{2}v}{d{t}^2}=6\left(2-\frac{2}{\sqrt{3}}\right)-12 \\ \Rightarrow \frac{-12}{\sqrt{3}}<0 \\ \text{So, velocity is maximum at}\mathit{\text{t}}=\left(2-\frac{2}{\sqrt{3}}\right). \\ \mathrm{Again}, \\ \frac{da}{dt}=6t-12 \\ \mathrm{For}\mathrm{maximum}\mathrm{or}\mathrm{minimum}\mathrm{values}\mathrm{of}\mathrm{a},\mathrm{we}\mathrm{must}\mathrm{have} \\ \frac{da}{dt}=0 \\ \Rightarrow 6t-12=0 \\ \Rightarrow t=2 \\ \mathrm{Now}, \\ \frac{d^{2}a}{d{t}^2}=6>0 \\ \text{So, acceleration is minimum at}\mathit{\text{t}}\text{=2.} \end{gathered}$$