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