Drag on a ball - NewtonII & calculus

[Dan27]

Homework Statement

A football of mass 0.4 kg travels horizontally above the ground. The ball experiences
a drag force of D_T = 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?

Cheers

 

[HallsofIvy]

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.

 

[Dan27]

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!