Question

A particle moves along a straight line and its velocity depends on time as $v=6t-3t^2$ where $v$ is in $\text{m/s}$ and $t$ in seconds. Find the average speed for first 4 seconds. (in $\text{m/s}$)

Answer 1

For average speed $=\frac{\text{Total distance}}{\text{Total time}}$
Velocity, $v=6t-3t^2=3t(2-t)$
i.e., $v\ge 0$ for $0\le t\le 2$
and $v\le 0$ for $2<t\le 4$
Thus, average speed
$=\frac{\left|{\int }_0^{2}vdt\right|+\left|{\int }_2^{4}vdt\right|}{{\int }_0^{4}dt}$
$=\frac{|6{\int }_0^{2}tdt-3{\int }_0^2{t}^{2}dt|+|6{\int }_2^{4}tdt-3{\int }_2^4{t}^{2}dt|}{4}$
$=\frac{\left|4\right|+|-20|}{4}$
$=\frac{24}{4}=6\text{m/sec}$