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The distance travelled by a particle in ...

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

Text Solution

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Distance travelled during time 0 to 5.0 s is `s=(2.5m//s^2) (5.0)^2 = 62.5m " Average speed during this time " = (62.5)/5 = 12.5m//s`
`V_(av) = 12.5 m//s`
b) `s=(2.5 m//s^2)t^2`
`(ds)/(dt) = 2.5m//s^2 xx2xxt=5xxt`
`v=5.0 xx 5.0=25m//s`
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