Question

Let the instantaneous velocity of a rocket just after launching be given by the expression $v=2t+3t^2$ (where $v$ is in $m/s$ and $t$ is in seconds). Find out the distance travelled between $t=2s$ to $t=3s$ (in $m$).

• A. 36
• B. 24
• C. 27
• D. 12

Answer 1

The correct option is B 24

To find the distance travelled, we need to integrate $v$ as $v$ always remains positive after $t=0s$. The limits of integration will be from $2s$ to $3s$ as we have to find the distance travelled between $t=2s$ to $t=3s$.
$x={\int }_2^{3}vdt={\int }_2^3\left(2t+3t^2\right)dt\Rightarrow x={\left(\frac{2t^2}{2}+\frac{3t^3}{3}\right|}_2^3={\left(t^2+t^3\right|}_2^3\Rightarrow x=24m$