A Force Depending on Velocity and Displacement

  • #1

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 :)
 

Answers and Replies

  • #2
Charles Link
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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.
 
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  • #3
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
 
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  • #4
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.
 
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  • #5
Charles Link
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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 ##.
 
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