Question

If motion of a particle is represented by $s=e^t+4^t-\cos t$, 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. $e^t-4^t\ln 4+\cos t$
• B. $e^t+4^t(\ln 4{)}^2+\cos t$
• C. $e^t+4^t(\ln 4{)}^2-\cos t$
• D. $e^t+4^t\ln 4+\cos t$

Answer 1

The correct option is B $e^t+4^t(\ln 4{)}^2+\cos t$

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, $S=e^t+4^t-\cos t$
So, we have $\frac{ds}{dt}=\frac{d(e^t+4^t-\cos t)}{dt}$
$\Rightarrow \frac{d{e}^t}{dt}+\frac{d{4}^t}{dt}-\frac{d\cos t}{dt}=e^t+4^t\ln 4+\sin t$
Thus, $\frac{d^{2}s}{d{t}^2}=\frac{d(e^t+4^t\ln 4+\sin t)}{dt}$
$\Rightarrow \frac{d{e}^t}{dt}+\ln 4\frac{d{4}^t}{dt}+\frac{d\sin t}{dt}=e^t+4^t(\ln 4{)}^2+\cos t$