# Homework Help: Intro to DIFF EQ (help )

1. Feb 11, 2006

### A_I_

intro to DIFF EQ (help!!)

a drag chute must be designed to reduce the speed of a 3000-lb dragster from 220 mph to 50 mph in 4 sec. Assume that the drag force is proportional to the velocity.

a) what value of the drag coefficient k is needed to accomplish this?
b) how far will the dragster travel in the 4-sec interval?

the way i reasonned is the following:

m(dv/dt) = sum of all forces
m(dv/dt) = -mg -kv (F=-kv drag force "opposite direction to the velocity).

mv' +kv = -mg
v' +(k/m)v = -g

integrating factor = e^(k/m)t

multiplying the integrating factor by the diff eq and integrating

we get: v(t) = -(mg/k) +Ce^(-k/m)t

initial conditions t=0 v=v(0)=220

--> C = v(0) +(mg/k)

thus v(t) = -(mg/k)(1 - e^(-k/m)t) +v(0)e^(-k/m)t

now in order to find the value of k
we have v(t)=50 mph
v(0) = 220 mph
g =32
m =3000
t = 4 sec

but we have a k^-1 and a k in the exponential.
how can we get the value of k?

it's a sophomore level course.
I don't need very complicated equations :).

Thanks,
Joe

2. Feb 11, 2006

### Fermat

Your dragster is moving horizontally, so you won't need an mg term - that acts in the vertical direction anyway, and m(dv/dt) and kv act in the horizontal.

That should simplify it a bit.

Also, remember to keep your units compatible. If you are using imperial measure, then your velocity shoud be in feet/sec, not mph.

3. Feb 11, 2006

### A_I_

i dont know how to transform from mph to feet per sec :)
can you help me? :)

4. Feb 11, 2006

### Fermat

There are 1760 yards in a mile and 3 ft in a yard
Convert miles to feet and divide by 3600 (60*60) to bring it from mph to ft/sec

5. Feb 11, 2006

### A_I_

i got for 50 mph = 73.3 ft/sec
and for 220 mph = 322.6 ft/sec

my solution was v(t) = v(0)e^(-k/m)t

v(t) = 73.3
v(0) = 322.6
t = 4 sec
m = 3000 lb

i got a value of k =1111.1
but in solutionary we have k = 0.2469

i must have done a terrible mistake.
can u tell me what is it?

thanks

6. Feb 11, 2006

### Fermat

I've gone over the problem a few times, and I can't justify the given solution, of k = 0.2964.
I get the same result as yourself, k = 1111.2.
It looks like a wrong answer in the book/solution set.

7. Feb 11, 2006

### A_I_

well i'm sorry to make you work it few times.
And in fact, 1111.3 is the right answer.

I still have some problem in the second part of the problem

mv' = -kv

m (dv/dt) = -kv

m (dv/dx)(dx/dt) = -k(dx/dt)

m(dv/dx) = -k

mdv = -kdx

integating mv = -kx
thus we end up with a negative value for x
or is there another way to solve for x?

thanks again

8. Feb 11, 2006

### Fermat

That's the correct way to solve for x.
x is supposed to be negative so that you have a decreasing exponential function. One that starts at 220 mph and ends up at 50 mph.

Oh, btw, You didn't have to convert the velocity from mph to ft/s. Not in this particular problem, that is. You would have got the same value for k if you had used mph, since it only depended on the ratio of the velocities.

9. Feb 11, 2006

### HallsofIvy

Notice that 3000 pounds is not mass. It is weight which is the gravitational force on the car. Since F= ma so m= F/a= 3000/32.2 slugs (or poundals).