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

$\dfrac{dv(t)}{dt} = a(t) \Rightarrow dv(t) = a(t)dt$

Integrating this expression, we get

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

$\:\:\:\:\:\:\:= \frac{1}{3}t^3 - 2t^2 + 5t + C_1$

Also, recall that

$\dfrac{ds(t)}{dt} = v(t)$ or

$\displaystyle s(t) = \int v(t)dt = \int (\frac{1}{3}t^3 - 2t^2 + 5t + C_1)dt$

$\:\:\:\:\:\:\:= \frac{1}{12}t^4 - \frac{2}{3}t^3 + \frac{5}{2}t^2 + C_1t + C_2$

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

$s(1) = \frac{1}{12}(1)^4 - \frac{2}{3}(1)^3 + \frac{5}{2}(1)^2 + C_1(1)$

$= \frac{1}{12} - \frac{2}{3} + \frac{5}{2} + C_1 = \frac{23}{12} + C_1 = 20$

or

$C_1 = \dfrac{217}{12}$

Therefore, s(t) can be written as

$s(t) = \frac{1}{12}t^4 - \frac{2}{3}t^3 + \frac{5}{2}t^2 + \frac{217}{12}t$