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

Explanation

a. The distance travelled during the 0 to 5.0 is
`s=(2.5m/s^2)(5.0)^2=62.5m`.
The average speed during this time is
`v_(av)=(62.5m)/(5s)=12.5m/s`
b. `s=(2.5 m/s^2)t^2`
or, `(ds)/(dt)=(2.5m/s^2)(2t)=(5.0 m/s^2)t`.
At t=5.0 s the speed is
`v=(ds)/(dt)=(5.0 m/s^2)(5.0s)=25m/s`.