Constant Acceleration Question

1. Sep 6, 2006

rogerhughston

The question is

A Train pulls away froma station with a constant acceleration of .4m/s^2. A passenger arrives at the track 6.0s after the end of the train has passed the very same point. what is the slowest constant speed at which she can run and catch the train?

I am rather confused on what to do but i have some idea

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distance of train in x(6s) = 1/2 * (.4m/s^2) * (6s)^2 = 7.2m
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So the person is 7.2m away when she gets there

But how do create 2 equations and set them equal to eachother?
Im thinking it might be as simple as expressing the train as a parabolic equation which intercepts at 7.2, and the linear line starts at 0, but how do i express accelleration in an equation that i can graph??

2. Sep 7, 2006

tony873004

How fast is the train moving by the time the end of the train reaches that point? Off the top of my head, it seems like not enough info given.

Yes, you'll need simultaneous equations and set them equal to each other.

3. Sep 7, 2006

HallsofIvy

Staff Emeritus
Yes, with constant acceleration .4 m/s2, initial speed 0, the train will have gone .2 t2 m in t seconds. Running with constant speed v, a person will run vt m in t seconds. But you have to be careful about where you "start measuring" time. You have to decide whether to take t= 0 when the train starts or when the person starts running- there is a 6 seconds difference. Either choice works.

1) t= 0 when train starts moving. After t seconds, the train will have gone 0.2t2 m. Since the person doesn't start running until 6 seconds later, at time t> 6, the person will have been running for t- 6 seconds. The person will have gone v(t- 6) m. In order to catch the train, we must have 0.2t2= v(t- 6). You could, for example, use the quadratic formula to solve that for t, depending on the parameter v. What is the smallest v for which that equation has a solution for t?

2. t= 0 when the person starts running. At time t, the person will have gone vt meters. At time t,the train will have been accelerating for t+ 6 seconds and will have gone 0.2(t+ 6)2 meters. We must have
0.2(t+6)2= vt. Again, that's a quadratic equation for t with parameter v. What's the smallest v for which that equation has a solution? Try it both ways and see if you get the same answer.