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## Homework Statement

"As two trains move along a track, their conductors suddenly notice that the are headed toward each other. Figure 2-28 gives their velocities as functions of time

*t*as the conductors slow th trains. The figure's vertical scaling is set by v

_{s}= 40.0 m/s. The slowing processes begin when the trains are 200 m apart. What is their separation when both trains have stopped?"

## Homework Equations

x - x

_{0}= v

_{0}t + ½at

^{2}

v

_{0A}= 40 m/s

a

_{A}= -0.8 m/s

^{2}

v

_{0B}= -30 m/s

a

_{A}= 0.75 m/s

^{2}

*Answer from textbook: 40 m*

## The Attempt at a Solution

First, I listed all the equations for the movement of the trains.

##v_A(t) = -\frac{4}{5}t+40##

##s_A(t) = -\frac{2}{5}t^2+40t##

##v_B(t) = \frac{3}{4}t-30##

##s_B(t) = \frac{3}{8}t^2-30t##

Plugging in the appropriate times...

##s_A(5) = -10+200 = 190##

##s_B(4) = 6-120 = -114##

Adding them together, I do not get the answer listed in the book.

##s_A(5) + s_B(4) = 76 m ≠ 40m##

Though there are two very peculiar things I noticed with the velocity equations. When I set them to zero:

##0 = -\frac{4}{5}t+40##

##-40 = -\frac{4}{5}t##

##-40 = -\frac{4}{5}t##

##t = 50 s##

##0 = \frac{3}{4}t-30##

##30 = \frac{3}{4}t##

##t = 40s##

Those are clearly not the values depicted in Figure 2-28. I may be wrong, but these velocity equations may have two different constants. Anyway, the second thing; when I set both displacements to (4), I somehow end up with the correct answer, despite the fact that train A doesn't stop at 4 seconds.

##s_A(4) = -6.4+160 = 153.6##

##s_B(4) = 6-120 = -114##

Adding them together (not subtracting like in the previous operation),

##s_A(4) + s_B(4) = 39.6 = 40 m##

Overall, I'm very confused on how to do such a simple problem, and at this point, I'm just messing with numbers until I get the correct answer. I feel it's very inefficient, and there must be a better way of learning. Anyway, I don't understand this problem at all.