The distance travelled by a particle in time t is given by `s=(2.5)t^(2)`.Find (a)the average speed of the particle during the time 0 to 5.0s, and (b)the instantanous speed at t=5.0s. Here s is in metres.

Explanation

(a)The distance travelled during time 0 to 5.0s is=(2.5)`(5.0)^(2)`=62.5 m.
The average speed during this time is `v_(av)=(62.5m)/(5s)=12.5m//s`
(b)s=`(2.5)t^(2)` or, `(ds)/(dt)=(2.5)(2t)`=(5.0)t.
At t=5.0s the speed is `V=(ds)/(dt)=(5.0)(5.0)=2m//s`