Find the position of the particle. a(t) = 2t + 3, s(0) = 8, v(0) = −3A particle is moving with the given data. Find the position of the particle. a(t) = 2t + 3, s(0) = 8, v(0) = −3
1 Answer
beckden | High School Teacher | (Level 1) Educator
`(d(v(t)))/(dt) = a(t)` , `(d(s(t)))/(dt) = v(t)`
So v(t) is the antiderivative of a(t) and s(t) is the antiderivative of v(t).
`v(t) = t^2 + 3t + C` , since `v(0) = -3 = 0^2 + 3(0) + C` , so `C = -3`
`v(t) = t^2 + 3t - 3`
`s(t) = t^3/3 + 3/2t^2 - 3t + C` , since `s(0) = 8 = 0^3/3 + 3/2(0)^2 - 3(0) + C` , `C = 8`
so my final answer is
`s(t) = t^3/3 + 3/2t^2 - 3t + 8`