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Second Order ODE

I can't seem to finish this second order ODE question. I'll type out the entire question word for word, because I think it's my interpretation of the initial conditions that are throwing me off.

Two cars travel along a road. The driver of the second car brakes if the car in front is travelling
more slowly, and speeds up if the car in front is travelling faster. If U and V are the speeds of
the two cars (V corresponding to the car in front), and if c is a positive constant, which of the
following ODE models summarises the actions of the driver of the second car?

(a) U"= c(U - V ), (b) U"= c(V - U), (c) V"= c(U - V ), (d) V"= c(V - U).

The front car travels at a constant speed 60kph (kilometres/hour). The second car starts at
40kph a distance of 20m behind the ?rst car. After a long time the distance between the two
cars is 30m.
Find c

What I've done so far:

Selected b) and I'm pretty certain about that.
Redefined U as x'(t)
Chosen my initial conditions to be x(0) = 20 and x'(0) = 40.
I arrive at the result:

x(t) = \displaystyle20\left(\frac{e^{-ct} - 1}{c} + 3t + 1\right)

However my problem arises when using the "when t is a large value, the distance is 30".
When this equation tends to infinity, it gets a little crazy if I follow the operation of dividing through by t and whacking infinity in.
Thus, I can't find c.

Hints on what to do next would be preferred over a straight out answer, but both appreciated. :smile:

(edited 13 years ago)

Reply 1

ttoby

Think about what x(t) is representing. Is it the distance that the car at the rear has travelled? In that case, at time 0 you would say that the car at the rear is in position 0. Hence x(0)=0. I agree with your x'(0)=40

So this gives you U'=c(V-U) hence x''=c(60-x'). I'm assuming that where you wrote U''=c(V-U) you meant U'=c(V-U).

Then for the car in front, the position would be given as 20 + 60t. So you would want to find 20+60t - x as t tends to infinity.