$\mathrm{s}=3{\mathrm{t}}^2–8\mathrm{t}+5$ is the distance reached by a body in $\mathrm{t}$ seconds, and the body will stop after

The correct option is C

4 / 3 seconds

Step: 1 Given data:

$\mathrm{s}=3{\mathrm{t}}^2–8\mathrm{t}+5$

Step: 2 Formula used:

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

Step: 3 Calculate the time when the body will stop:

$\mathrm{s}=3{\mathrm{t}}^2–8\mathrm{t}+5$

$\mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}=\frac{\mathrm{d}\left(3{\mathrm{t}}^2-8\mathrm{t}+5\right)}{\mathrm{dt}}=6\mathrm{t}-8$

For the time when the body will stop, the velocity = 0

$$\begin{gathered} 6\mathrm{t}-8=0 \\ 6\mathrm{t}=8 \\ \mathrm{t}=\frac{8}{6}\mathrm{or}\frac{4}{3} \end{gathered}$$

Hence, option C is correct.