Question

The position $x$ of a particle varies with time as $x=a{t}^2-b{t}^3$. The acceleration of a particle is zero at time $T$equal to

• A. $\frac{2a}{3b}$
• B. $\frac{a}{b}$
• C. $\frac{a}{3b}$
• D. $0$

Answer 1

The correct option is C

$\frac{a}{3b}$

Step 1:Given:

Acceleration at time $T$, $a=0$.

Displacement equation, $x=a{t}^2-b{t}^3$

Step 2:Write an expression of accleration:

$a=\frac{d^{2}x}{d{t}^2}$

Step 3:Differentiate two times $x$ with respect to $t$ to get the accleration:

$$\begin{gathered} \frac{dx}{dt}=2at-3b{t}^2 \\ \frac{d^{2}x}{d{t}^2}=2a-6bt \end{gathered}$$

Step 4:Substitute $0$ for $\frac{d^{2}x}{d{t}^2}$ and $T$ for $t$ in above acceleration expression:

$$\begin{gathered} 2a-6bT=0 \\ T=\frac{2a}{6b} \\ T=\frac{a}{3b} \end{gathered}$$

Hence,Option(C) is correct.