A Force Depending on Velocity and Displacement

[Bill_Nye_Fan]

Homework Statement

A body of mass 1 kg moves so that at time $t$ seconds, its displacement is $x$ meters from a fixed origin. The body is acted upon by a force of $\frac{\sqrt{x-1}}{v^2}$ Newtons, where $v$ is the velocity in m/s. The body is initially at rest 1 meter North of the origin. Express $x$ in terms of $t$ and hence find the point in time when the displacement and acceleration have the same magnitude. Express all answers correct to 3 significant figures.

Homework Equations

Using calculus to convert between displacement, velocity and acceleration
$a=\frac{dv}{dt}=\frac{d^2x}{dt^2}=v\cdot \frac{dv}{dx}$
$v=\frac{dx}{dt}$
$a=f\left(x\right),\ \frac{d^2x}{dt^2}=f\left(x\right),\ \left(\frac{dx}{dt}\right)^2=2\int _{ }^{ }f\left(x\right)dx,\ v^2=2\int _{ }^{ }f\left(x\right)dx$

The Attempt at a Solution

$a=\frac{\sqrt{x-1}}{v^2}$, $v^2=\frac{\sqrt{x-1}}{a}$ (equation 1)
Using the last rule in the relevant equations:
$v^2=2\int _{ }^{ }\frac{\sqrt{x-1}}{v^2}dx$ (equation 2)
Substituting the first equation into the second:
$\frac{\sqrt{x-1}}{a}=2\int _{ }^{ }\frac{\sqrt{x-1}}{\left(\frac{\sqrt{x-1}}{a}\right)}dx=2\int _{ }^{ }adx$
$\int _{ }^{ }adx=\frac{\sqrt{x-1}}{2a}$
$a=\frac{d\left(\frac{\sqrt{x-1}}{2a}\right)}{dx}$
$a=\frac{\left(\frac{a}{\sqrt{x-1}}-2\sqrt{x-1}\cdot \frac{da}{dx}\right)}{4a^2}$
$4a^3=\frac{a}{\sqrt{x-1}}-2\sqrt{x-1}\cdot \frac{da}{dx}$
$-2\sqrt{x-1}\cdot \frac{da}{dx}=4a^3=\frac{a}{\sqrt{x-1}}$
$\frac{da}{dx}=\frac{a}{2\left(x-1\right)}-\frac{2a^3}{\sqrt{x-1}}$

From here I'm not sure how to simplify this further. Any help is greatly appreciated :)

 

[Charles Link]

I believe I solved it, and the approach is much different. I won't give you the answer, but I will lead you to it. Start with $(\frac{d^2x}{dt^2})(\frac{dx}{dt})^2=\sqrt{x-1}$. Look for a simple trial function for $x=x(t)$ that will simplify the right side. The obvious choice is $x(t)=t^n +1$. Try to solve for $n$ and see what you get. This one is not the answer yet=but start with that, and see if you can get close.

 

[Bill_Nye_Fan]

I believe I solved it, and the approach is much different. I won't give you the answer, but I will lead you to it. Start with $(\frac{d^2x}{dt^2})(\frac{dx}{dt})^2=\sqrt{x-1}$. Look for a simple trial function for $x=x(t)$ that will simplify the right side. The obvious choice is $x(t)=t^n +1$. Try to solve for $n$ and see what you get. This one is not the answer yet=but start with that, and see if you can get close.

Thanks for the advice :) I'll start working through some trial functions

 

[Bill_Nye_Fan]

Update: I just noticed that the formula $a=v\cdot \frac{dv}{dx}$ works a lot better in this scenario. Almost all the previous questions I was doing were using $a=2\int _{ }^{ }f\left(x\right)dx$ so I kind of got tunnel vision of only using that formula.

$a=\frac{\sqrt{x-1}}{v^2}$
$v\cdot \frac{dv}{dx}=\frac{\sqrt{x-1}}{v^2}$
$v^3\cdot dv=\sqrt{x-1}\cdot dx$
$\int _{ }^{ }v^3dv=\int _{ }^{ }\sqrt{x-1}dx$
$\frac{v^4}{4}=\frac{2}{3}\left(x-1\right)^{\frac{3}{2}}+c$

"The body is initially at rest 1 meter North of the origin"
$x=1, v=0$
Therefore, $c=0$

$\frac{v^4}{4}=\frac{2}{3}\left(x-1\right)^{\frac{3}{2}}$
$v=\left(\frac{8}{3}\left(x-1\right)^{\frac{3}{2}}\right)^{\frac{1}{4}}$
$\frac{dx}{dt}=\left(\frac{8}{3}\left(x-1\right)^{\frac{3}{2}}\right)^{\frac{1}{4}}$
$\frac{dt}{dx}=\frac{1}{\left(\frac{8}{3}\left(x-1\right)^{\frac{3}{2}}\right)^{\frac{1}{4}}}$
$t=\int _{ }^{ }\frac{1}{\left(\frac{8}{3}\left(x-1\right)^{\frac{3}{2}}\right)^{\frac{1}{4}}}dx$
$t=\frac{4\cdot 6^{\frac{1}{4}}\left(x-1\right)}{5\left(x-1\right)^{\frac{3}{8}}}+c$

Initially the particle is 1 m from the origin: $t=0, x=0$. Therefore, $c=0$

$t=\frac{4\cdot 6^{\frac{1}{4}}\left(x-1\right)}{5\left(x-1\right)^{\frac{3}{8}}}$
$\frac{5t}{4\cdot 6^{\frac{1}{4}}}=\left(x-1\right)^{\frac{5}{8}}$

$x\left(t\right)=\left(\frac{5t}{4\cdot 6^{\frac{1}{4}}}\right)^{\frac{8}{5}}+1$

$v(t)=\frac{dx}{dt}=\left(\frac{5^{\frac{3}{5}}}{2^{\frac{3}{5}}\cdot 3^{\frac{2}{5}}}\right)t^{\frac{3}{5}}$

$a(t)=\frac{dv}{dt}=\frac{\left(\frac{3}{2}\right)^{\frac{3}{5}}}{5^{\frac{2}{5}}}t^{-\frac{2}{5}}$

Then to find the point in time when the magnitudes are equal we just solve $x(t)=a(t)$ which gave me $t=0.291$ (which was the answer in the book)

So this wasn't actually that difficult of a question once I actually started using the right formula haha.

 

[Charles Link]

Very good ! :) :) $\\$ In my way of solving it, ultimately the solution is to let $x=ct^n+1$. You then solve for $n$ by setting the powers of $t$ equal on both sides of the equation, and likewise for $c$.