`a(t)=36t+8` The acceleration is the derivative with respect to time of the velocity, so

`v(t)=18t^2+8t+C` Since v(0)=10 we have `10=18(0)+8(0)+C==>C=10` thus `v(t)=18t^2+8t+10`

The velocity is the derivative with respect to time of the distance function, so

`s(t)=6t^3+4t^2+10t+C` Since s(0)=14 we have:

`s(t)=6t^3+4t^2+10t+14`

For t=8 we get

`s(8)=6(8^3)+4(8^2)+10(8)+14=3422`

** The position of...**

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

`a(t)=36t+8` The acceleration is the derivative with respect to time of the velocity, so

`v(t)=18t^2+8t+C` Since v(0)=10 we have `10=18(0)+8(0)+C==>C=10` thus `v(t)=18t^2+8t+10`

The velocity is the derivative with respect to time of the distance function, so

`s(t)=6t^3+4t^2+10t+C` Since s(0)=14 we have:

`s(t)=6t^3+4t^2+10t+14`

For t=8 we get

`s(8)=6(8^3)+4(8^2)+10(8)+14=3422`

**The position of the particle is 3422 units.**