Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 enter image description here

Help me out, Appreciate your help. Thank.

share|improve this question
add comment

2 Answers

up vote 3 down vote accepted

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

share|improve this answer
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.