Question

The position x of a particle varies with time t as $x = at^{2} - bt^{3}.$ The acceleration of the particle will be zero at time t equal to

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

Answer 1

Answer: (C)

$\frac{a}{3b}$

Answer 2

Given $x = at^{2} - bt^{3}$ ∴ velocity $(v) = \frac{dx}{dt} = 2at - 3bt^{2}$ and acceleration (1) =$\frac{dv}{dt} = 2a - 6bt.$

When acceleration = 0 ⇒ $2a - 6bt = 0$⇒ $t = \frac{2a}{6b} = \frac{a}{3b}$.