Question

Find the average velocity of the position function $s\left(t\right)=3t^2-6t$ on the interval from $t=0\text{sec}$ to $t=4\text{sec}$.

• A. $4\text{m/s}$
• B. $6\text{m/s}$
• C. $3\text{m/s}$
• D. $5\text{m/s}$

Answer 1

The correct option is B $6\text{m/s}$

Given $s\left(t\right)=3t^2-6t$
So, at $t=0,s\left(t\right)=0$, it means particle has started its motion from origin, so, we will check whether it has returned to origin or not.
$\Rightarrow s\left(t\right)=0\Rightarrow 3t^2-6t=0$
$\Rightarrow 3t(t-2)=0\Rightarrow t=0,2\text{sec}$
Thus, the particle will return to origin at $t=2\text{sec}$, afterwards it will not return to origin.
So, the displacement of particle from $t=2\text{sec}$ to $t=4\text{sec}$ will be
$s=3[t^2{]}_2^4-6[t{]}_2^4=3(16-4)-6(4-2)=36-12=24\text{m}$
Hence, average velocity is $\overline{v}=\frac{\text{Total displacement}}{\text{Total time}}=\frac{24}{4}=6\text{m/s}$