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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 ast t=5.0 is

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(a) The distance travelled during time 0 to 5.0 s is,
`S = (2.5)(5.0)^(2) = 62.5` m
The average speed during this time is
`v_("avg")= ("distance travelled")/("time taken") = (62.5)/(5) = 12.5 m//s`
(b) Given `S = (2.5)t^(2)`
Intantaneous speed = `(dS)/(dt) = (d)/(dt) (2.5t^(2)) = (2.5)(2t) = 5t`
`therefore ` At `t = 5.0s` the speed is
`[v]_(t=5S)= (5.0) xx (5.0) = 25 m//s`
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