Question

Find the average velocity of the position function $s\left(t\right)=2t^2-4t$ on the interval from $t=0\text{sec}$ to $t=5\text{sec}$.

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

Answer 1

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

Given $s\left(t\right)=2t^2-4t$
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
$\Rightarrow s\left(t\right)=0\Rightarrow 2t^2-4t=0$
$\Rightarrow 2t(t-2)=0\Rightarrow t=0,2\text{sec}$
Thus, the particle will return to origin at $t=2\text{sec}$, afterwards it will continue its motion.
So, the displacement of particle from $t=2\text{sec}$ to $t=5\text{sec}$ will be
$s=2[t^2{]}_2^5-4[t{]}_2^5=2(25-4)-4(5-2)=42-12=30\text{m}$
Hence, average velocity is $\overline{v}=\frac{\text{Total displacement}}{\text{Total time}}=\frac{30}{5}=6\text{m/s}$