Question

If motion of a particle is represented by $s=e^t+4t^3-\cos t$, then the acceleration of the particle at any time $t$ is
[Hint: Acceleration of the particle is $\frac{d^{2}s}{d{t}^2}$]

• A. $e^t+12t^2+\sin t$
• B. $e^t+24t-\cos t$
• C. $e^t+24t+\cos t$
• D. $e^t+12t^2-\sin t$

Answer 1

The correct option is C $e^t+24t+\cos t$

As we know that the acceleration of the particle at any time $t$ is given by
$\frac{d^{2}s}{d{t}^2}\Rightarrow \frac{d\left(\frac{ds}{dt}\right)}{dt}$
Given, $S=e^t+4t^3-\cos t$
So, we have $\frac{ds}{dt}=\frac{d(e^t+4t^3-\cos t)}{dt}$
$\Rightarrow \frac{d{e}^t}{dt}+4\frac{d{t}^3}{dt}-\frac{d\cos t}{dt}=e^t+12t^2+\sin t$
Thus, $\frac{d^{2}s}{d{t}^2}=\frac{d(e^t+12t^2+\sin t)}{dt}$
$\Rightarrow \frac{d{e}^t}{dt}+12\frac{d{t}^2}{dt}+\frac{d\sin t}{dt}=e^t+24t+\cos t$