Ordinary Differential Equations

Question

A rocket sled having an initial speed of 191 mi/h is slowed by a channel of water. Assume that during the braking process, the acceleration a is given by a(v)=−μυ^2, where v is the velocity and μ is a constant. Find the time τ required to slow the sled to 18 mi/h as a function of μ.

 

So far I have dv/dt = -μv^2. I then solved it as a separable to get:

1/μv = t+C

 

Not sure what to do from here.

Answer 1

You are on the right track.  The next step is to use the given initial condition to complete the solution, and then solve for the time that makes v = 18 mph.

 

dv/dt = -μv²

 

Separate the variables.

 

dv/v² = -μdt

 

Integrate both sides.

 

-1/v(t) = -μt + c

 

-1/v(0) = -μ(0) + c = c

 

c = -1/v(0)

 

-1/v(t) = -μt - 1/v(0)

 

v(t) = 1/[μt + 1/v(0)]

 

v(t) = v(0) / [μv(0)t + 1]

 

You were given v(0) = 191 mph, so set v(τ) = 18 mph.

 

18 = 191/(191μτ + 1)

 

Solve for τ in terms of μ.  You should be able to finish from here.