The accelearation of an object is given by $a(t)=5t^\frac{2}{3}+2e^{-t}$. Find $s(t)$ Suppose that the acceleration of an object is given by $a(t)=5t^\frac{2}{3}+2e^{-t}$. The object's initial velocity, $v(0)$, is $11$ and the object's initial position, $s(0)$,is $-5$.Find $s(t)$.
Even by reading this question, make me dizzy.I have no clue where to start it from and how to solve.
However I manage to differentiate it.
$$a(t)=5t^\frac{2}{3}+2e^{-t}$$
$$\frac{d}{dt}=\frac{10}{3t^\frac{1}{3}}-2e^{-t}$$
Here is the question photo with multiple choice answer

Help me out, Appreciate your help.
Thank.
 A: Your equation $a(t)=5t^\frac{2}{3}+2e^{-t}$ is a 2nd order differential of the displacement, $s(t)$.
The fundamentals of displacement-velocity-acceleration relationsship is, given displacement relative to time of an object $s(t)$
$$
\begin{align*}
\frac{d}{dt}s(t) &= v(t)\\
\frac{d}{dt}v(t) &= a(t) = \frac{d^2s}{dt^2}
\end{align*}
$$
So, in your question, given $a(t)$ and being asked to find $s(t)$, you should integrate $a(t)$ twice instead of differentiating it.
So
$$
\begin{align*}
v(t) &= \int a(t) \,dt
\\&= \int 5t^{2/3} + 2e^{-t} \,dt
\\&= 3\int \frac 5 3 t^{2/3}\,dt -2\int -e^{-t} \,dt
\\&= 3t^{5/3} -2e^{-t} + C
\end{align*}
$$
Solving for $v(t)$,
$$
\begin{align*}
v(0) &= 3(0)^{5/3} -2e^{(0)} + C
\\11&= C-2\\
C &= 13\\
\therefore v(t) &= 3t^{5/3} -2e^{-t} + 13
\end{align*}
$$
And,
$$
\begin{align*}
s(t) &= \int v(t) \,dt
\\&= \int 3t^{5/3} -2e^{-t} + 13 \,dt
\\&= \frac 98\int \frac  83 t^{5/3} +2\int -e^{-t} + \int 13 \,dt
\\&= \frac 98 t^{8/3} + 2e^{-t} + 13t + D
\end{align*}
$$
Solving for $s(t)$,
$$
\begin{align*}
s(0) &= 0 + 2e^{0} + D\\
-5 &= 2 + D\\
D &= -7\\
\therefore s(t) &= \frac 98 t^{8/3} + 2e^{-t} + 13t -7
\end{align*}
$$
And this matches choice B
A: Just integrate twice: $$v(t)=v(0)+\int_0^ta(\tau) d\tau $$ $$s(t)=s(0)+\int_0^t v(\tau)d \tau $$
