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 $m{s}^{-1},t$ is in $s$. The average velocity of the object between $t=0$ and $t=2s$ is

• A. $0$
• B. $3m{s}^{-1}$
• C. $2m{s}^{-1}$
• D. $4m{s}^{-1}$

Answer 1

Answer: (C)

Given, velocity, $v=6t-3t^2$
As we know that,
$v=\frac{dx}{dt}$
Here, $x$ is the displacement of the particle.
Now, $dx=vdt$
Integrate on the both sides, limit $t=0$ to $t=2$, we get
$\therefore x=\underset{0}{\overset{2}{\int }}vdt=\underset{0}{\overset{2}{\int }}\left(6t-3t^2\right)dt$
$={\left[\frac{6t^2}{2}\right]}_0^2-{\left[\frac{3t^3}{3}\right]}_0^2={\left[3t^2\right]}_0^2-{\left[t^3\right]}_0^2$
$=\left[3(2{)}^2-3(0{)}^2\right]-\left[(2{)}^3-(0{)}^2\right]$
$=[12-0]-[8-0]=12-8=4m$
Average velocity, $v_{\text{avg }}=\frac{\text{ Total displacement }}{\text{ Total time taken }}$ $=\frac{4}{2}=2m/s$
Hence, the average velocity of the object between
$t=0$ to $t=2s$ is $2m/s$