Drag on a ball - NewtonII & calculus

AI Thread Summary
The discussion focuses on solving a physics problem involving a football experiencing a drag force proportional to the square of its speed. The drag force equation is set up using Newton's second law, leading to a differential equation that requires integration. Participants clarify the steps for separating variables and integrating to find the time it takes for the ball's speed to decrease from 15 m/s to 12 m/s. There is some confusion regarding the treatment of negative signs in the equations, but the approach to solving the differential equation is affirmed. The conversation emphasizes the importance of understanding the relationship between drag force, mass, and acceleration in this context.
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Homework Statement



A football of mass 0.4 kg travels horizontally above the ground. The ball experiences
a drag force of DT = 0.01v2 where v is instantaneous speed.
Immediately after the kick the ball has a speed of 15 m/s.
How long does it take for the speed to drop to 12 m/s?

Homework Equations



F = ma

0.01v2 = ma

0.01v2 = m(dv/dt)

The Attempt at a Solution



I realize that the acceleration is not constant and so Newton II will have to be used in creating a differential equation that can be integrated. But I'm struggling to take it from the differential with confidence in what I'm doing.

Can someone please let me know if I'm following the right lines and how to progress? o:)

Cheers
 
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Your equation is -0.01v^2= .4\frac{dv}{dt} (".4" because you are given that m= .4 kg, "-" because drag is always opposite the direction of motion.).

You can "separate" that by multiplying both sides by -100dt and dividing both sides by v^2 to get dt= -40v^2 dv. Now integrate both sides.
 
HallsofIvy said:
Your equation is -0.01v^2= .4\frac{dv}{dt} (".4" because you are given that m= .4 kg, "-" because drag is always opposite the direction of motion.).

You can "separate" that by multiplying both sides by -100dt and dividing both sides by v^2 to get dt= -40v^2 dv. Now integrate both sides.

That's brilliant thank you :)

I just assumed that I could neglect the negative by taking the acceleration or \frac{dv}{dt} to also be negative?

Also, by dividing by v^2 would the RHS not be \frac{-40}{v^2} dv

Thank you!
 
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