Q71 / 150 · MHT CET 2008 · Mathematics
The velocity of a particle at time $t$ is given by the relation $v=6 t-\frac{t^{2}}{6}$. The distance traveled in $3 \mathrm{~s}$ is, if $s=0$ at $t=0$
- A$\frac{39}{2}$
- B$\frac{57}{2}$
- Correct answerC$\frac{51}{2}$
- D$\frac{33}{2}$
Solution
Given, $v=\frac{d s}{d t}=6 t-\frac{t^{2}}{6}$
On integrating both sides, we get
$$
s=3 t^{2}-\frac{t^{3}}{18}+\text { constant }
$$
Now, put $s=0$ at $t=0$, we get constant $=0$
$$
\therefore \quad s=3 t^{2}-\frac{t^{3}}{18}
$$
Now, distance traveled in $3 s=3(3)^{2}-\frac{(3)^{3}}{18}$
$$
=27-\frac{27}{18}=\frac{51}{2}
$$
Hence, the correct answer is option: 'C'.