Question

Let the instantaneous velocity of a rocket, just after landing, is given by the expression, $\vartheta =2t+3t^2$ (where $\vartheta$ is in $m{s}^{-1}$ and $t$ is in seconds.). Find out the distance travvelled by the rocket from $t=2s$ to $t=3s$.

• A. $24$
• B. $14$
• C. $8$
• D. $48$

Answer 1

Answer: (A)

$24$

Given,

$v=2t+3t^2$

The distance travelled by the rocket from $t=2sec$ to $t=3sec$

$S={\int }_2^{3}vdt$

$S={\int }_2^3(2t+3t^2)dt$

$S=[\frac{2t^2}{2}+\frac{3t^3}{3}{]}_2^3$

$S=[t^2+t^3{]}_2^3$

$S=(3\times 3+3\times 3\times 3)-(2\times 2+2\times 2\times 2)$

$S=36-12=24m$

The correct option is A.