Equation of displacement for any particle is $s = 3t^3 + 7t^2 + 14t + 8$ m. Its acceleration at time $t = 1$ s

Question

Equation of displacement for any particle is $s = 3t^3 + 7t^2 + 14t + 8$ m. Its acceleration at time $t = 1$ sec is:

Accepted Answer

1.  **Velocity ($v$):** Instantaneous velocity is the rate of change of displacement, $v = \frac{ds}{dt}$ .
    Given $s = 3t^3 + 7t^2 + 14t + 8$.
    Differentiating with respect to $t$:
    $v = \frac{d}{dt}(3t^3 + 7t^2 + 14t + 8) = 9t^2 + 14t + 14$.
2.  **Acceleration ($a$):** Instantaneous acceleration is the rate of change of velocity, $a = \frac{dv}{dt}$ .
    Differentiating $v$ with respect to $t$:
    $a = \frac{d}{dt}(9t^2 + 14t + 14) = 18t + 14$.
3.  **Calculation at $t = 1$ sec:**
    Substitute $t = 1$ into the acceleration equation:
    $a = 18(1) + 14 = 32 	ext{ m/s}^2$.

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