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The distance travelled by a particle in ...

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

Text Solution

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a. The distance travelled during the 0 to 5.0 is
`s=(2.5m?s^2)(5.0)62=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`.
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