How Does a Skydiver's Velocity Change with Distance After Parachute Deployment?

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Will Freedic
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Homework Statement


A sky driver reaches his terminal velocity of 373 ms^-1 and deploys his parachute at a height of 1500m, the drag force =bv^2 and weigh can be ignored. Find an expression for velocity as a function of distance traveled from the point the parachute was deployed

The Attempt at a Solution


I can find velocity as a function of time easy enough but not a function of distance[/B]
 
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Will Freedic said:

Homework Statement


A sky driver reaches his terminal velocity of 373 ms^-1 and deploys his parachute at a height of 1500m, the drag force =bv^2 and weigh can be ignored. Find an expression for velocity as a function of distance traveled from the point the parachute was deployed
A terminal velocity of 373 m/s is quite a feat. It's also greater than Mach 1, which makes it unlikely to occur. Are you sure a decimal point hasn't been misplaced?
 
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There is a standard trick for eliminating time from an acceleration equation.
##a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}##.
Note that if we apply this to F=ma and integrate wrt x we get ##\int F.dx=\frac 12 mv^2##.
 
Thank you!, so is it correct to say that:

Cv2 = ma
cv2 = mv dvdx.
Cm/x = v2 / 2 + c
Cheers
 
Will Freedic said:
Thank you!, so is it correct to say that:

Cv2 = ma
cv2 = mv dvdx.
Cm/x = v2 / 2 + c
Cheers
A couple of problems there. First, what happened to gravity? Secondly, I don't understand what you did to get from cv2 = mv dvdx (presumably you meant mv dv/dx) to Cm/x = v2 / 2 + c
 
haruspex said:
A couple of problems there. First, what happened to gravity? Secondly, I don't understand what you did to get from cv2 = mv dvdx (presumably you meant mv dv/dx) to Cm/x = v2 / 2 + c
Thank you for your help, i am new to these forums so am a little slow!, managed to get there now tho