A particle moves along a straight line such that its displacement at any time $t$ is given by $S = t^3 - 6t^2

Question

A particle moves along a straight line such that its displacement at any time $t$ is given by $S = t^3 - 6t^2 + 3t + 4$ metres. The velocity when the acceleration is zero is:

Accepted Answer

1.  **Velocity ($v$):** Instantaneous velocity is the rate of change of displacement with respect to time, $v = \frac{dS}{dt}$ .
    Given $S = t^3 - 6t^2 + 3t + 4$.
    Differentiating with respect to $t$:
    $v = \frac{d}{dt}(t^3 - 6t^2 + 3t + 4) = 3t^2 - 12t + 3$.
2.  **Acceleration ($a$):** Instantaneous acceleration is the rate of change of velocity, $a = \frac{dv}{dt}$ .
    Differentiating $v$ with respect to $t$:
    $a = \frac{d}{dt}(3t^2 - 12t + 3) = 6t - 12$.
3.  **Condition for Zero Acceleration:** Set $a = 0$.
    $6t - 12 = 0 \implies 6t = 12 \implies t = 2 	ext{ s}$.
4.  **Calculate Velocity:** Substitute $t = 2 	ext{ s}$ into the velocity equation.
    $v = 3(2)^2 - 12(2) + 3$
    $v = 3(4) - 24 + 3$
    $v = 12 - 24 + 3 = -9 	ext{ ms}^{-1}$.

Step-by-Step Solution

Practice All PHYSICS Questions