Question

The velocity of an object moving in a straight line path is given as a function of time by $v = 6t - 3t^2$. where $v$ is in $ms^{-1} , t $ is in $s$. The average velocity of the object between $t = 0$ and $t = 2s$ is

• A. $0$
• B. $3 \; ms^{-1}$
• C. $ 2 \; ms^{-1}$
• D. $4 \; ms^{-1}$

Answer 1

The Correct Option is C

Given, velocity, $v=6 t-3 t^{2}$
As we know that,
$v=\frac{d x}{d t}$
Here, $x$ is the displacement of the particle.
Now, $d x=v d t$
Integrate on the both sides, limit $t=0$ to $t=2$, we get
$\therefore x=\int\limits_{0}^{2} v d t=\int\limits_{0}^{2}\left(6 t-3 t^{2}\right) d t $
$=\left[\frac{6 t^{2}}{2}\right]_{0}^{2}-\left[\frac{3 t^{3}}{3}\right]_{0}^{2}=\left[3 t^{2}\right]_{0}^{2}-\left[t^{3}\right]_{0}^{2} $
$=\left[3(2)^{2}-3(0)^{2}\right]-\left[(2)^{3}-(0)^{2}\right] $
$=[12-0]-[8-0]=12-8=4 \,m$
Average velocity, $v_{\text {avg }}=\frac{\text { Total displacement }}{\text { Total time taken }}$ $=\frac{4}{2}=2 m / s$
Hence, the average velocity of the object between
$t=0$ to $t=2 s$ is $2 m / s$