Question

The figure shows the graph of velocity-time for a particle moving in a straight line. If the average speed for $6sec$ is ' $b$ ' and the average acceleration from $0sec$ to $4sec$ is ' $c^{\prime }$ find magnitude of bc (in $m^2/s^3$ ).

Answer 1

Answer: 25

Average speed $=\frac{\text{ dis tan ce travelled }}{\text{ time taken }}$
$b=\frac{\text{ total area }}{\text{ total time }}$
$=\frac{10+20}{6}=\frac{30}{6}=5m/s$
Average acceleration $=\frac{\text{ change in velocity }}{\text{ time taken }}$
$C=\frac{10-(-10)}{4}=\frac{20}{4}=5m/s^2$
$bc=(5)(5)=25m^2/s^3$