Question

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

Options

•  \[\frac{16}{9} \text { unit }/\sec\]

• \[- \frac{32}{3} \text { unit }/\sec\]

• \[\frac{4}{3} \text { unit }/\sec\]

• \[- \frac{16}{3} \text { unit }/\sec\]

Answer 1

 \[- \frac{16}{3} \text { unit }/\sec\]

\[\text { According to the question,}\]

\[s = t^3 - 4 t^2 + 5\]

\[ \Rightarrow \frac{ds}{dt} = 3 t^2 - 8t\]

\[ \Rightarrow \frac{d^2 s}{d t^2} = 6t - 8\]

\[ \Rightarrow 6t - 8 = 0 \left[ \text { As velocity deminishes, then }\frac{d^2 s}{d t^2}=0 \right]\]

\[ \Rightarrow t = \frac{4}{3}\]

\[\text { Now }, \left( \frac{ds}{dt} \right)_{t = \frac{4}{3}} = 3 \left( \frac{4}{3} \right)^2 - 8\left( \frac{4}{3} \right)\]

\[ \Rightarrow \frac{ds}{dt} = \frac{16}{3} - \frac{32}{3}\]

\[ \Rightarrow \frac{ds}{dt} = - \frac{16}{3}\text { unit }/\sec\]