`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.
2 Answers
gsarora17 | (Level 2) Associate Educator
`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 | (Level 3) Honors
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 `