A particle is moving in a straight line its displacement at any instant t is given by `x=5t^(2)+20t^(3)`. Find the average acceleration in the interval t = 0 to t = 3 seconds.

Explanation

Given equation `x=5t^(2)+20t^(3)`
Time `t=0, " to "t=3s.`
Velocity of particle `v=(dx)/(dt)`
`v=(d)/(dt)(5t^(2)+20t^(3))`
Differentiate with respect to 't'
`v=10t+60t^(2)`
Again differentiate with respect to 't'
`v=10+120t`
t=0.
Again differentiate with repect to 't'
`v=10+120t`
t=0,
`v_(0)=10+120(0)`
`v_(0)=10m//s`
`"at t"=2sec, v_(2)=10+120(t)`
`v_(2)=10+120(2)`
`v_(2)=10+240.`
`v_(2)=250ms^(-1)`
`Deltav=v_(2)-v_(0)`
`=250-10`
`=240ms^(-1)`
`Deltat=t_(2)-t_(1)=2-0=2`
The average acceleration is,
`a_("ave")=(Deltav)/(Deltat)=(240)/(2)`
`a_("ave")=120ms^(-2)`