Velocity as a function of distance problem

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SUMMARY

The discussion focuses on solving the problem of determining the force F(x), position x(t), and force F(t) for a ball with mass 'm' whose velocity varies with distance as v(x) = ax-n. Participants clarified that acceleration a can be expressed as a = dv/dt and further manipulated to relate distance and time through integration. The key steps involve integrating the equation xndx = a dt to find x in terms of constants a, n, and time t, ultimately leading to the expression for force F(t) based on mass and acceleration.

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Homework Statement



The velocity of a ball of mass 'm' varies with the distance v(x) = ax-n, where a and n are positive constants.

Determine F(x), x(t) and F(t).

Homework Equations


F = ma
a = dv/dt
v = dx/dt


The Attempt at a Solution


If velocity were the function of time, i would have done it with no problem. I determined the acceleration, which is -nax-n-1.
For v, i integrated 'a' with respect to t. But velocity's dependence in distance confused me. Any help will be highly appreciated.
 
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There is a simple way for these.
a=dv/dt. This can be written as a=dx/dt*dv/dx.
 
v = dx/dt = a*x^-n
So x^n*dx = a*dt.
Find the integration and then find x in terms of a, n and t.
Then d^2(x)/dt^2 will give acceleration in terms of a,n and t. from that you can find F(t)
Acceleration a = dv/dt = dv/dx*dx/dt = dv/dx*v. Then F(x) = ma
 
Last edited:
To rl.bhat

Thanks. I tried to follow your instruction.
Integration of x^n*dx = a*dt gave me (1/n+1)x^(n+1) = v. I didn't find x in terms of a, n and t.
 
(1/n+1)x^(n+1) = v.
This should be
(1/n+1)x^(n+1) = at
x^(n+1) =a(n+1)*t
x = ...?
 

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