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.

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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

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Just integrate twice: $$v(t)=v(0)+\int_0^ta(\tau) d\tau$$ $$s(t)=s(0)+\int_0^t v(\tau)d \tau$$