Question

The distance travelled by a particle in time $t$ is given by $s=\left(2.5m/s^2\right)t^2$ . Find
(a) the average speed of the particle during the time $0$ to $5.0s$ , and
(b) the instantaneous speed at $t=5.0s.$

Answer 1

Distance travelled is given as,

$s=\left(2.5m/s^2\right)t^2$

$=2.5\times 5^2$

$=62.5m$

Average speed is given as,

$v_{av}=\frac{62.5}{5}$

$=12.5m/s$

Instantaneous speed is given as,

$v_i=\frac{ds}{dt}$

$=\left(2.5m/s^2\right)\times 2t$

$=2.5\times 2\times 5$

$=25m/s$