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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
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