# Train with constant acceleration, catching it with constant velocity?

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1. Oct 7, 2014

### timnswede

1. The problem statement, all variables and given/known data
A train pulls away from a station with a constant acceleration of 0.4 m/s2. A passenger arrives at a point next to the track 6 s after the end of the train has passed the very same point. What is the slowest constant speed at which she can run and still catch the train? (changed 6.4 s to 6 s and .41 m/s2 to .4)

2. Relevant equations
V=V0+at
X=X0+V0t+.5at2

3. The attempt at a solution
So I got the distance and the velocity of the train when the passenger shows up which is 7.2+2.4t+.2t2.
For the speed of the passenger I got V(t-6). I set them equal to each other and got V(t-6)=7.2+2.4t+.2t2 I am not sure where to go after this. I saw some other forums with the same questions, but I didn't really understand what to do, any help or hints as to what to do now would be appreciated.

Last edited: Oct 7, 2014
2. Oct 7, 2014

### Staff: Mentor

Can you show how you arrived at this? That doesn't look like a distance and a velocity to me (looks more like an equation of motion), and the numbers don't seem to fit your problem statement.
The problem statement says that the passenger arrives 6.4 seconds after the train departs. But you've used 6. That can't lead to a good ending...

3. Oct 7, 2014

### timnswede

Oh sorry I copied the problem down wrong :)
Well I got the velocity after 6 seconds from V=V+at, and then then the position from the velocity and the time. To be completely honest I'm not sure if I am going about this right.

4. Oct 7, 2014

### Staff: Mentor

Okay, I see that you've edited the original problem statement in your first post. (you should leave a note in the post to say that you've made changes when significant things like the data values are altered. Just saying...).

So, given these new values I can see where you've arrived at an equation of motion for the train beginning at the instant that the passenger arrives. That equation is your:

d = 7.2+2.4t+.2t2

That equation describes the train's position from that moment when the passenger arrives at the platform. So whatever the passenger does from that moment, that's when the clock starts ticking for that equation and whatever equation you cook up for the passenger. No need to bias the passenger's time by 6 seconds since it's already incorporated into the equation of motion for the train.

Can you proceed from here?

5. Oct 7, 2014

### timnswede

So then I would have Vpt=7.2+2.4t+.2t2 right? How would I go about solving that? Or am I missing something obvious here?

6. Oct 7, 2014

### Staff: Mentor

Nope, you're missing something imaginary

If you solve for t, that is, the time the passenger and train meet, then you will be solving a quadratic equation for the time that they meet, right? If her speed happened to be less than what was required, what might happen to the solutions of that quadratic?

7. Oct 7, 2014

### timnswede

Then it would be imaginary right? So I am doing something wrong :/
Still lost as to where to go from after I get 7.2+2.4t+.2t2.

8. Oct 7, 2014

### Staff: Mentor

No, not doing something wrong. The solution will only be real if the passenger's speed is sufficient. If the time that the passenger and train meet is imaginary then they don't meet in reality.

So, how do you tell if the solution of a quadratic will be real or imaginary"?

9. Oct 7, 2014

### timnswede

If there is a negative number under the square root. But I feel like there should be an answer, since without considering human physical limits the passenger should be able to run as fast as needed to catch the train.

10. Oct 8, 2014

### Staff: Mentor

Correct. Do you agree that so long as the passenger's speed is great enough that the discriminant will be a positive value? And that if her speed is too low it will be negative (signalling that she cannot catch the train)? What can you say about the case where the discriminant is zero?

11. Oct 8, 2014

### timnswede

If it were zero then she would catch the train at the slowest speed needed? I think I figured it out. I subtracted the Vt (passengers position) from the position function of the train and got 0=7.2+(2.4-v)t+.2t2. Then took the discriminant and set it equal to zero so, and ended up with (2.4-v)2-4(.2)(7.2)=0. Solved that for V and got an answer of 4.8 m/s which I believe is right. Does that look correct to you?

12. Oct 8, 2014

### Staff: Mentor

That looks good

13. Oct 8, 2014

### timnswede

Great, thank you for your help!