Question

The velocity of a particle at time $t$ is given by the relation $v=6t-\frac{t^2}{6}$ . The distance traveled in $3s$ 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, $v=\frac{ds}{dt}=6t-\frac{t^2}{6}$
On integrating both sides, we get
$s=3t^2-\frac{t^3}{18}+$constant
Now, put $s=0$ at $t=0,$ we get constant $=0$
$\therefore s=3t^2-\frac{t^3}{18}$
Now, distance traveled in $3s=3(3{)}^2-\frac{(3{)}^3}{18}$
$=27-\frac{27}{18}=\frac{51}{2}$