Question

If $\mathrm{s}={\mathrm{ae}}^{\mathrm{t}}+(\mathrm{b}/{\mathrm{e}}^{\mathrm{t}})$is the distance in seconds described by a particle in $\mathrm{t}$ seconds, then the particle's acceleration at the time $\mathrm{t}$ is

• A. proportional to t
• B. Proportional to s
• C. s
• D. Constant

Answer 1

The correct option is C

s

Step: 1 Given Data:

$\mathrm{s}={\mathrm{ae}}^{\mathrm{t}}+(\mathrm{b}/{\mathrm{e}}^{\mathrm{t}})$

Step: 2 Formula used:

The velocity of the particle $\left(\mathrm{v}\right)=\frac{\mathrm{ds}}{\mathrm{dt}}$

The acceleration of the particle $=\frac{\mathrm{dv}}{\mathrm{dt}}$

Step: 3 Calculate acceleration:

$\mathrm{s}={\mathrm{ae}}^{\mathrm{t}}+(\mathrm{b}/{\mathrm{e}}^{\mathrm{t}})$

$$\begin{gathered} \mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}=\frac{\mathrm{d}\left({\mathrm{ae}}^{\mathrm{t}}+(\mathrm{b}/{\mathrm{e}}^{\mathrm{t}})\right)}{\mathrm{dt}} \\ ={\mathrm{ae}}^{\mathrm{t}}–{\mathrm{be}}^{-\mathrm{t}} \end{gathered}$$

Acceleration = $\frac{\mathrm{d}\left({\mathrm{ae}}^{\mathrm{t}}–{\mathrm{be}}^{-\mathrm{t}}\right)}{\mathrm{dt}}={\mathrm{ae}}^{\mathrm{t}}–{\mathrm{be}}^{-\mathrm{t}}(-1)={\mathrm{ae}}^{\mathrm{t}}+{\mathrm{be}}^{-\mathrm{t}}$

${\mathrm{ae}}^{\mathrm{t}}+{\mathrm{be}}^{-\mathrm{t}}=\mathrm{s}$

Acceleration $=\mathrm{s}$

Hence, option C is correct.