Question

$\mathrm{s}=\sqrt{({\mathrm{at}}^2+2\mathrm{bt}+\mathrm{c})}$ is the distance traveled by a particle traveling in a straight line at the time $\mathrm{t}$. The particle is accelerated

• A. proportional to t
• B. proportional to s
• C. proportional to s^-3
• D. none of these

Answer 1

The correct option is C

proportional to s^-3

Step: 1 Given data:

$\mathrm{s}=\sqrt{({\mathrm{at}}^2+2\mathrm{bt}+\mathrm{c})}$

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:

$\begin{array}{rcl}\mathrm{s}& =& \sqrt{({\mathrm{at}}^2+2\mathrm{bt}+\mathrm{c})}\\ \mathrm{v}& =& \frac{\mathrm{ds}}{\mathrm{dt}}\\ & =& \frac{\mathrm{d}\left(\sqrt{{\mathrm{at}}^2+2\mathrm{bt}+\mathrm{c}}\right)}{\mathrm{dt}}\\ & =& \frac{2\mathrm{at}+2\mathrm{b}}{2\mathrm{s}}\end{array}$

Step 4: Calculating acceleration:

Acceleration = $\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{d}\left({\displaystyle \frac{2\mathrm{at}+2\mathrm{b}}{2\mathrm{s}}}\right)}{\mathrm{dt}}=\left(\frac{{\mathrm{s}}^2\mathrm{a}-{\left(\mathrm{at}+\mathrm{b}\right)}^2}{{\mathrm{s}}^3}\right)$

Since Acceleration is directly proportional to ${\mathrm{s}}^{-3}$.

Hence option C is correct.