# Finding the velocity of a block down an incline plane after in moves a distance $x_0$

Consider a block sliding down an incline plane at an angle $\theta$ with the horizontal. For the acceleration as a function of $\theta$ I find $$\ddot{x}=g \ \sin\theta$$ My text then claims we can find the block's velocity after it moves a distance $x_0$ from rest by multiplying both sides by $2\dot{x}$ and doing the following:

$$2\dot{x}\ddot{x}=2\dot{x}g \ \sin \theta$$ $$\frac{d}{dt}(\dot{x}^2)=2g \sin\theta\frac{dx}{dt}$$ $$\int_0^{v_0^2}d(\dot{x}^2)=2g\sin\theta\int_0^{x_0}dx$$ $${v_0}^2=2g\sin\theta \ x_0$$ $$v_0=\sqrt{2g\sin\theta \ x_0}$$

I think I understand up until the 3rd line. The $dt$'s disappear because both sides are exact differentials, yes? Then in the next step, why does $\dot{x}$ (the velocity) vary from 0 to ${v_0}^2$? Thanks in advance.

Also, is there another way to do this?

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If $\left.\dot{x}\right|_{x=x_0}=v_0$ then $\left.\dot{x}^2\right|_{x=x_0}=v_0^2$, no? –  Kyle Kanos Oct 8 '13 at 13:00
I will fill in the missing step. In the second line, he has $\frac{d}{dt}(\dot{x}^2)=2g \sin\theta\frac{dx}{dt}$, then integrates w.r.t. $t$ to get $\int_{t_i}^{t_f}\frac{d}{dt}(\dot{x}^2) dt=2g\sin\theta\int_{t_i}^{t_f}\frac{dx}{dt} dt$. Then he does $u$-substitution on both sides. On the left $u$ is $\dot{x}^2$ and on the right $u$ is $x$. This gives you his third line. If anyone wants to turn this into a complete answer, go ahead. –  NowIGetToLearnWhatAHeadIs Nov 8 '13 at 0:09

A simpler way to this is to use the conservation of energy. If surface is frictionless, Then P.E = K.E ($mg x_o sin \theta= 1/2 m v_o^2$), this would lead to the same final answer).

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When you go from

$\frac{d}{dt}(\dot{x}^2)=2g \sin\theta\frac{dx}{dt}$

to the next step, the rigorous way to think about it is to integrate wrt $t$,

$\int_{t_0}^{t_1}\!dt\, \frac{df}{dt} = \int_{f(t_0)}^{f(t_1)} df$.

In this case, $f(t) = \dot x$, and $f(t_1) = v_0$

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