(adsbygoogle = window.adsbygoogle || []).push({}); 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

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Intro to DIFF EQ (help )

**Physics Forums | Science Articles, Homework Help, Discussion**