`a(t) = t^2 - 4t + 6, s(0) = 0, s(1) = 20` A particle is moving with the given data. Find the position of the particle.

Textbook Question

Chapter 4, 4.9 - Problem 64 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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`a(t)=t^2-4t+6`

`a(t)=v'(t)`

`v(t)=inta(t)dt`

`v(t)=int(t^2-4t+6)dt`

`v(t)=t^3/3-4(t^2/2)+6t+c_1`

`v(t)=t^3/3-2t^2+6t+c_1`

`s(t)=intv(t)dt`

`s(t)=int(t^3/3)-2t^2+6t+c_1)dt`

`s(t)=(1/3)(t^4/4)-2t^3/3+6t^2/2+c_1t+c_2`

`s(t)=t^4/12-(2t^3)/3+3t^2+c_1t+c_2`

Let's find constants c_1 and c_2 . given s(0)=0 and s(1)=20

`s(0)=0=c_2`

`s(1)=20=1/12-2/3+3+c_1`

`20=(1-8+36)/12+c_1`

`20=29/12+c_1`

`c_1=20-29/12=211/12`

`s(t)=t^4/12-2t^3/3+3t^2+211/12t`

scisser's profile pic

scisser | (Level 3) Honors

Posted on

Integrating for velocity:

`v(t) = (1/3)t^3 - 2t² + 6t + C `

Integrating for position:

`s(t) = (1/12)t^4 - (2/3)t^3 + 3t² + Ct + D `

Using the 2nd given value

`20 = (1/12) - (2/3) + 3 + C + D `

Solving D

`D = -C - 17.58 `

Back to this equation:
`s(t) = (1/12)t^4 - (2/3)t^3 + 3t² + Ct + D `

`s(t) = (1/12)t^4 - (2/3)t^3 + 3t² + Ct - C - 17.58 `

Using the first given value:

`0 = 0 - 0 + 0 + 0 - C - 17.58 `
`C = -17.58 `

Back this equation again:
`s(t) = (1/12)t^4 - (2/3)t^3 + 3t² + Ct + D `

`s(t) = s(t) = (1/12)t^4 - (2/3)t^3 + 3t² - 17.58t - C - 17.58 `

`s(t) = (1/12)t^4 - (2/3)t^3 + 3t² - 17.58t + 17.58 - 17.58 `

`s(t) = (1/12)t^4 - (2/3)t^3 + 3t² - 17.58t `

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