Question

The position $x$ of a body is defined by equation $x=P{t}^2-Q{t}^3$. The acceleration of the particle will be zero at time equal to

• A. $\frac{P}{Q}$
• B. $\frac{Q}{P}$
• C. $\frac{P}{3Q}$
• D. $\frac{3P}{Q}$

Answer 1

Answer: (C)

$\frac{P}{3Q}$

Given,

$x=P{t}^2-Q{t}^3$

velocity, $v=\frac{dx}{dt}=2Pt-3Q{t}^2$

Acceleration, $a=\frac{dv}{dt}=2P-6Qt$

The acceleration of the particle will be zero at time t'

$\frac{d^{2}x}{d{t}^2}=0$

$2P-6Q{t}^{\prime }=0$

$2P=6Q{t}^{\prime }$

$t^{\prime }=\frac{P}{3Q}$

The correct option is C.