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 seconds 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

The correct option is C $\frac{51}{2}$

$\frac{ds}{dt}=6t-\frac{t^2}{6}$
Now on integrating both sides $s=3t^2-\frac{t^2}{18}+$constant,
(where s is distance)
Now put t=0, then s=0 gives constant equal to 0 and putting t=3, we get
$s=3\left(3\right)62-\frac{3^3}{18}=27-\frac{27}{18}=\frac{51}{2}.$
Aliter :${\int }_0^{s}ds={\int }_0^3(6t-\frac{t^2}{6})dt=\frac{51}{2}$