Question

The velocity of a particle at time $ t $ is given by the relation $ v=6t-\frac{t^{2}}{6} $ . The distance traveled in $ 3 \,s $ is, if $ s = 0 $ at $ t = 0 $

• A. $ \frac{39}{2} $
• B. $ \frac{57}{2} $
• C. $ \frac{51}{2} $
• D. $ \frac{33}{2} $

Answer 1

Answer: (C)

Given, $v=\frac{d s}{d t}=6 t-\frac{t^{2}}{6}$
On integrating both sides, we get
$s=3 t^{2}-\frac{t^{3}}{18}+$constant
Now, put $s=0$ at $t=0,$ we get constant $=0$
$\therefore \,\, s=3 t^{2}-\frac{t^{3}}{18}$
Now, distance traveled in $3 s=3(3)^{2}-\frac{(3)^{3}}{18}$
$=27-\frac{27}{18}=\frac{51}{2}$