Acceleration to velocity with coordinate $x$

Question

Can someone please help me in finding the formula used to get the answer to this question?

An object moves along $x$-axis. In any coordinate $x$, the acceleration is $a=x^4$ (SI units). If the object goes from rest in $x=1$m, what velocity it will get at $x=2$m? \begin{align*} A)\quad 4.59 \textrm{ m/s}\\ B)\quad 3.52 \textrm{ m/s}\\ C)\quad 2.47 \textrm{ m/s}\\ D)\quad 5.66 \textrm{ m/s}\\ E)\quad 1.41 \textrm{ m/s}\\ \end{align*}

Answer B is the correct answer

Answer 1

If $v=\frac{dx}{dt}$ is the velocity at time $t$, we have $$\frac{dv}{dt}=x^4.$$ There is a standard trick for solving this kind of DE. Multiply both sides by $v$. We get $$v\frac{dv}{dt}=vx^4=x^4\frac{dx}{dt}.$$ Now we can integrate with repect to $t$. On the left, we get $\frac{1}{2}v^2$, and on the right we get $\frac{1}{5}x^5 +C$. Thus $$\frac{1}{2}v^2=\frac{1}{5}x^5 +C.$$ Use your initial condition to find that $C=-\frac{1}{5}$. Now we know $v^2$ at position $x$.

Answer 2

Hint: Think about the relationship between velocity and acceleration in terms of calculus.

Let $v(x)$ be the velocity at position $x$; $a(x)$ be the acceleration at position $x$:

$$\int a(x)\ dx=v(s)+C,\ a(x)=v(x)\frac{d}{dx}$$