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 instantaneous speed at `t=5.0s` Here s is in metres.

Explanation

The distance travelled during tiem `0"to"5.0s` is `=(2.5)(5.0)^(2)=62.5m.`
the average speed during this time is `v_(av)=(62.5m)/(5s)=12.5m//s`
`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)=25m//s`