# At what distance will the particle be?

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A particle is moving on the x-axis with the effect of the friction $F=-kx^2$ but at the moment $t=0$ it is at the position $x=0$ and it has velocity $x'=4$.

If $k=1/4, m=1$ at what distance will it be at $t=2$ ?

Do we use Newton' law $\sum F=0$ ?

My Attempt:
We have that $F=mv\frac{dv}{dx} ⇒−kx^2=mv\frac{dv}{dx}$.

asked Nov 2, 2016
retagged Nov 2, 2016
Use the fact that $F=m \frac{dv}{dt} = mv \frac{dv}{dx}$

## 1 Answer

2 votes

Best answer

Using $F = ma$, and given $m =1$, we have $a(x) = F = -kx^2$.

Now we can write $a(x)$ as,

$$a(x)=\frac{d v(x)}{d t}$$
and using the chain rule this becomes,

$$a(x) = \frac{d v(x)}{d x}\frac{d x}{d t}=\frac{d v(x)}{d x}v(x)$$

Thus,

$$a(x)dx = v(x)dv$$

So $$\int{-kx^2} dx = \int v(x)dv$$

Therefore,

$$\frac{-kx^3}{3} - 0 = \frac{v^2 - v_0^2}{2}$$

And after re-arranging gives velocity as a function of position,

$$v^2 = v_0^2 + \frac{-2kx^3}{3}$$

Therefore $$v(x) =\sqrt{16+ \frac{-2kx^3}{3}}$$

Now, let's find $t(x)$ which is required to solve the problem. We will first find it`s derivative:

$$\frac{d t(x)}{d x}=\frac{1}{v(x)}$$

Again, split the differentials and integrate (using $k = \frac{1}{4}$):

$$t =\int dt=\int \frac{dx}{v(x)}=\int \frac{dx}{\sqrt{16+ \frac{-x^3}{6}}}$$

By solving the above integral one can obtain $t$ as a function of $x$ and re-arrange to find $x$ when $t=2$.

answered Nov 2, 2016 by (1,486 points)
selected Nov 2, 2016 by Einstein