Question

If s = t³ − 4t² + 5 describes the motion of a particle, then its velocity when the acceleration vanishes, is

(a) $\frac{16}{9}\mathrm{unit}/\sec$

(b) $-\frac{32}{3}\mathrm{unit}/\sec$

(c) $\frac{4}{3}\mathrm{unit}/\sec$

(d) $-\frac{16}{3}\mathrm{unit}/\sec$

Answer 1

(d) $-\frac{16}{3}\mathrm{unit}/\sec$

$$\begin{gathered} \text{According to the question,} \\ s=t^3-4t^2+5 \\ \Rightarrow \frac{ds}{dt}=3t^2-8t \\ \Rightarrow \frac{d^{2}s}{d{t}^2}=6t-8 \\ \Rightarrow 6t-8=0\left[\text{As velocity deminishes, then}\frac{d^{2}s}{d{t}^2}\text{=0}\right] \\ \Rightarrow t=\frac{4}{3} \\ \mathrm{Now},{\left(\frac{ds}{dt}\right)}_{t=\frac{4}{3}}=3{\left(\frac{4}{3}\right)}^2-8\left(\frac{4}{3}\right) \\ \Rightarrow \frac{ds}{dt}=\frac{16}{3}-\frac{32}{3} \\ \Rightarrow \frac{ds}{dt}=-\frac{16}{3}\mathrm{unit}/\sec \end{gathered}$$