The distance s described by a particle in t seconds is given by

Solution:

It is given that, the distance s described by a particle in $t$ seconds is given by $s=a e^{t}+\frac{b}{e^{t}}$.

The acceleration of the particle at time $t$ is given by $\frac{d^{2} s}{d t^{2}}$.

$s=a e^{t}+\frac{b}{e^{t}}$

Differentiating both sides with respect to $t$, we get

$\frac{d s}{d t}=\frac{d}{d t}\left(a e^{t}+\frac{b}{e^{t}}\right)$

$\Rightarrow \frac{d s}{d t}=\frac{d}{d t}\left(a e^{t}+b e^{-t}\right)$

$\Rightarrow \frac{d s}{d t}=a e^{t}+b e^{-t} \times(-1)$

$\Rightarrow \frac{d s}{d t}=a e^{t}-b e^{-t}$

Again differentiating both sides with respect to t, we get

$\frac{d}{d t}\left(\frac{d s}{d t}\right)=\frac{d}{d t}\left(a e^{t}-b e^{-t}\right)$

$\Rightarrow \frac{d^{2} s}{d t^{2}}=a e^{t}-b e^{-t} \times(-1)$

 

$\Rightarrow \frac{d^{2} s}{d t^{2}}=a e^{t}+\frac{b}{e^{t}}=s$

Thus, the acceleration of the particle at time $t$ is $a e^{t}+\frac{b}{e^{t}}$.

The distance $s$ described by a particle in $t$ seconds is given by $s=a e^{t}+\frac{b}{e^{t}}$. Then the acceleration of the particle at time $t$ is equal to $a e^{t}+\frac{b}{e^{t}}$