- #1

Jenny Physics

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- 4

## Homework Statement

A man arrives ##T## seconds late to a train station. He watches the last three cars of the train pass by.

The next-to-the-next-to-last car takes time ##t## to pass by the passenger, the next-to-last car takes time ##t'## to pass by and the last car takes time ##t''## to pass by.

Assume the train's acceleration ##a## is constant. What is ##T##?

## Homework Equations

Kinematics.

## The Attempt at a Solution

The train departs at ##t=0## (man's time). At the time ##t_{l}=T## that the man arrives at the train station, the speed of the train is ##v_{l}=aT##. If the length of a car is ##L## then at time ##t_{2}=t_{l}+t## we have ##L=\frac{1}{2}at^{2}+v_{l}t##, at time ##t_{3}=t_{l}+t+t'## we have ##2L=\frac{1}{2}a(t+t')^{2}+v_{l}(t+t')## and at time ##t_{4}## we have ##3L=\frac{1}{2}a(t+t'+t'')^{2}+v_{l}(t+t'+t'')##. Therefore ##at^{2}+2v_{l}t=\frac{1}{2}a(t+t')^{2}+v_{l}(t+t')## or ##at^{2}+2aTt=\frac{1}{2}a(t+t')^{2}+aT(t+t')##. Solving for ##T## we get

$$T=\frac{t^{'2}+2tt'-t^{2}}{2(t-t')}$$

If we use ##L## and ##3L## we get instead ##\frac{3}{2}at^{2}+3v_{l}t=\frac{1}{2}a(t+t'+t'')^{2}+v_{l}(t+t'+t'')## or ##3at^{2}+6aTt=a(t+t'+t'')^{2}+2aT(t+t'+t'')## which leads to

$$T=\frac{(t+t'+t'')^{2}-3t^{2}}{2(2t-t'-t'')}$$

We could then equal these two expressions for ##T## and get some relation between ##t,t',t''##. This seems right to me but is it?