Question

The position is a particle along a straight line is denoted by the equation $x=6+12t-t^3$. (Here $x$ is in meter and $t$ is in second). The magnitude of acceleration of the particle when it is at rest is:-

• A. $6m/s^2$
• B. $12m/s^2$
• C. $18m/s^2$
• D. $24m/s^2$

Answer 1

The correct option is B $12m/s^2$

$\left(B\right)$

$v=\frac{dx}{dt}$ , $a=\frac{dv}{dt}=\frac{d^{2}x}{d{t}^2}$

$x=6+12t-t^3$

$v=12-3t^2$

$a=-6t$

Particle is at rest when $v=0$

$12-3t^2=0$

$t=2s$

Acceleration at $t=2s$

$a=-6t$

$a=-6\times 2=-12$

Magnitude of acceleration $=12m/s^2$