A particle is moving with acceleration a(t)=36t+8. Its position at time t=0 is s(0)=14 and its velocity at time t=0 is v(0)=10. What is its position at time t=8?
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Eric Bizzell
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`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.
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