# problem involving differencial equation

• Apr 17th 2007, 07:09 PM
clockingly
problem involving differencial equation
I found this calc problem in a chapter on improper integration, and it is confusing me so any help would be appreciated!

A car is travelling at 55 mph. The driver sees a traffic jam ahead and hits the brakes. Brakes apply friction. The car's velocity satisfies the differential equation v' = -1080v miles per hour per hour. How far does the car go after the brake is applied?
• Apr 17th 2007, 07:27 PM
topsquark
Quote:

Originally Posted by clockingly
I found this calc problem in a chapter on improper integration, and it is confusing me so any help would be appreciated!

A car is travelling at 55 mph. The driver sees a traffic jam ahead and hits the brakes. Brakes apply friction. The car's velocity satisfies the differential equation v' = -1080v miles per hour per hour. How far does the car go after the brake is applied?

v' + 1080v = 0

This is a homogeneous equation and has a solution:
v(t) = Ae^{-1080t} + B

So
v'(t) = -1080Ae^{-1080t}

-1080Ae^{-1080t} + 1080Ae^{-1080t} + 1080B = 0

Thus B = 0 and v(t) = Ae^{-1080t}

v(0) = 55, thus A = 55.

v(t) = 55e^{-1080t} (in units of mph.)

Thus:
x(t) = Int[v(t'), 0, t] = Int[55e^{-1080t'}, 0, t]

x(t) = 55/(-1080) * e^{-1080t'} |_{t' = 0}^{t' = t}

x(t) = -(55/1080)*(e^{-1080t} - 1)
(Note: t is in hours and x(t) is in miles.)

Now, what is x(0)?
x(0) = 0.

So the distance traveled by the car is simply x(t). Now, when does the car stop? The answer is NEVER!! (Look at v(t).) So we need to evaluate x(t) as t --> (infinity). So the distance traveled by the car is:
d = lim[x(t), t --> (infinity)] = lim[-(55/1080)*(e^{-1080t} - 1), t --> (infinity)]

d = -(55/1080)*(0 - 1) = 55/1080 miles

-Dan
• Apr 17th 2007, 07:34 PM
ecMathGeek
Quote:

Originally Posted by topsquark
v' + 1080v = 0

This is a homogeneous equation and has a solution:
v(t) = Ae^{-1080t} + B

So
v'(t) = -1080Ae^{-1080t}

-1080Ae^{-1080t} + 1080Ae^{-1080t} + 1080B = 0

Thus B = 0 and v(t) = Ae^{-1080t}

v(0) = 55, thus A = 55.

v(t) = 55e^{-1080t} (in units of mph.)

Thus:
x(t) = Int[v(t'), 0, t] = Int[55e^{-1080t'}, 0, t]

x(t) = 55/(-1080) * e^{-1080t'} |_{t' = 0}^{t' = t}

x(t) = -(55/1080)*(e^{-1080t} - 1)
(Note: t is in hours and x(t) is in miles.)

Now, what is x(0)?
x(0) = 0.

So the distance traveled by the car is simply x(t). Now, when does the car stop? The answer is NEVER!! (Look at v(t).) So we need to evaluate x(t) as t --> (infinity). So the distance traveled by the car is:
d = lim[x(t), t --> (infinity)] = lim[-(55/1080)*(e^{-1080t} - 1), t --> (infinity)]

d = -(55/1080)*(0 - 1) = 55/1080 miles

-Dan

I'm glad you went through this. I solved it, and got that the car never stops, and so deleted my solution thinking it was wrong. I just didn't think to solve it as t -> infinity.