The position $x$ of a particle varies with time $t$ as $x = at^2 - bt^3$. The acceleration of the particle wil

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:

Accepted Answer

1.  **Velocity ($v$):** Instantaneous velocity is defined as the rate of change of position with respect to time, $v = \frac{dx}{dt}$ .
    Given $x = at^2 - bt^3$.
    Differentiating with respect to $t$: $v = \frac{d}{dt}(at^2 - bt^3) = 2at - 3bt^2$.
2.  **Acceleration ($a_{acc}$):** Instantaneous acceleration is defined as the rate of change of velocity with respect to time, $a_{acc} = \frac{dv}{dt}$ .
    Differentiating $v$ with respect to $t$: $a_{acc} = \frac{d}{dt}(2at - 3bt^2) = 2a - 6bt$.
3.  **Condition:** We need to find the time $t$ when the acceleration is zero.
    Set $a_{acc} = 0$:
    $2a - 6bt = 0$
    $6bt = 2a$
    $t = \frac{2a}{6b} = \frac{a}{3b}$.

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