Q87 / 125 · BITSAT 2010 · Physics
The position $\mathrm{x}$ of a particle varies with time $(\mathrm{t})$ as $\mathrm{x}=\mathrm{A} \mathrm{t}^{2}-\mathrm{B} \mathrm{t}^{3}$. The acceleration at time $\mathrm{t}$ of the particle will be equal to zero. What is the value of $t ?$
- A$\frac{2 \mathrm{~A}}{3 \mathrm{~B}}$
- B$\frac{\mathrm{A}}{\mathrm{B}}$
- Correct answerC$\frac{A}{3 B}$
- Dzero
Solution
Given that $x=A t^{2}-B t^{3}$
$$
\therefore \text { velocity }=\frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{~A} \mathrm{t}-3 \mathrm{~B} \mathrm{t}^{2}
$$
and acceleration
$$
=\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)=2 \mathrm{~A}-6 \mathrm{Bt}
$$
For acceleration to be zero $2 \mathrm{~A}-6 \mathrm{Bt}=0$.
$$
\therefore \quad \mathrm{t}=\frac{2 \mathrm{~A}}{6 \mathrm{~B}}=\frac{\mathrm{A}}{3 \mathrm{~B}}
$$
Hence, the correct answer is option: 'C'.