`a(t) = 3cos(t) - 2sin(t), s(0) = 0, v(0) = 4` A particle is moving with the given data. Find the position of the particle.

Textbook Question

Chapter 4, 4.9 - Problem 62 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to remember the relation between acceleration, velocity and position, such that:

`int a(t)dt = v(t) + c`

`int v(t) dx = s(t) + c`

You need to find first the velocity function, such that:

`int (3cos t - 2sint )dt = int 3cos t dt - int 2sin t dt`

`int (3cos t - 2sint )dt = 3sin t + 2cos t + c`

The problem provides the information that v(0) = 4, hence, you may find c:

`v(0) = 3sin 0 + 2cos 0 + c => 4 = 0 + 2 + c => c = 2`

Hence, the velocity function is `v(t) = 3sin t + 2cos t + 2`

You need to evaluate the position function, such that:

`int (3sin t + 2cos t + 2) dt = int 3sin t dt + int 2cos t dt + int 2dt`

`int (3sin t + 2cos t + 2) dt = -3cos t + 2sin t + 2t + c`

Hence, the position function `s(t) = -3cos t + 2sin t + 2t + c` can be completely determined, if the information provided by the problem is used, s(0) = 0.

`s(0) = -3cos 0 + 2sin 0 + 0 + c => 0 = -3 + c => c = 3.`

Hence, evaluating the position of the particle, yields `s(t) = -3cos t + 2sin t + 2t + 3.`

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scisser | (Level 3) Honors

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Integrate a(t) to get v(t)
`v(t) = 3 sin (t) + 2 cos (t) + C `

UseĀ  `v(0) = 4` to find C
`4 = 3 sin (0) + 2 cos (0) + C `
`C = 2 `

Thus,
`v(t) = 3 sin (t) + 2 cos (t) + 2 `

Integrate again: v(t) to get s(t)
`s(t) = -3 cos (t) + 2 sin (t) + 2t + c `

Use ` v(0) = 4` to find c
`0 = -3 cos (0) + 2 sin (0) + 2(0) + c `
`c = 3 `

Thus,
`s(t) = -3 cos (t) + 2 sin (t) + 2t + 3`

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