Question

A particle is moving with the given data. Find the position of the particle.

a(t) =
`t^(2)` − 4t + 5, s(0) = 0, s(1) = 20

How do I find s(t)=?

Answer 1

Recall that

`(dv(t))/(dt) = a(t) \Rightarrow dv(t) = a(t)dt`

Integrating this expression, we get

`\displaystyle v(t) = \int a(t)dt = \int(t^2 - 4t + 5)dt`

`\:\:\:\:\:\:\:= (1)/(3)t^3 - 2t^2 + 5t + C_1`

Also, recall that

`(ds(t))/(dt) = v(t)` or

`\displaystyle s(t) = \int v(t)dt = \int ((1)/(3)t^3 - 2t^2 + 5t + C_1)dt`

`\:\:\:\:\:\:\:= (1)/(12)t^4 - (2)/(3)t^3 + (5)/(2)t^2 + C_1t + C_2`

Next step is to find
`C_1\:\text{and}\:C_2`. We know that at t = 0, s = 0, which gives us
`C_2 = 0`. At t = 1, s = 20, which gives us

`s(1) = (1)/(12)(1)^4 - (2)/(3)(1)^3 + (5)/(2)(1)^2 + C_1(1)`

`= (1)/(12) - (2)/(3) + (5)/(2) + C_1 = (23)/(12) + C_1 = 20`

or

`C_1 = (217)/(12)`

Therefore, s(t) can be written as

`s(t) = (1)/(12)t^4 - (2)/(3)t^3 + (5)/(2)t^2 + (217)/(12)t`