Question

Given s(t)= -t^3+2t^2+(3/2) be the position of a particle moving along the x-axis at time t. At what time will the instantaneous velocity equal the average velocity over the time interval (0,4)?

Answer 1

Answer:
`(16)/(9)\ s`

Explanation:

Given

`s(t)=-t^3+2t^2+(3)/(2)`

Average velocity is given by

`v_(avg)=(\int vdt)/(\int dt)`

`v_(avg)=(-(t^4)/(4)+(2)/(3)t^3+1.5t)/(t)\\\\v_(avg)=-(t^3)/(4)+(2)/(3)t^2+1.5`

Now, equate the average and instantaneous velocity

`-(t^3)/(4)+(2)/(3)t^2+1.5=-t^3+2t^2+1.5\\\\\Rightarrow -(3t^3)/(4)+(4t^2)/(3)=0\\\\\Rightarrow t^2\left(-(3)/(4)t+(4)/(3)\right)=0\\\\\Rightarrow t=(16)/(9)\ s`