Kinematics One Dimensional help

[toesockshoe]

ugh, I am a moron too. btw, can you guys tell me what difficulty you would think this problem is? i would never have figured out the velocities equal each other if no one pointed it out... ugh. i feel like this is too hard a course for me.

 

[FallenLeibniz]

It's not exactly a hard one but not a piece of cake either. Don't worry. Textbooks are filled with a good amount of ones like these. The best thing to do would be to practice and/or study with a friend. You can do it. Just got to keep on truckin.

It may be helpful to keep units in your numbers just for awhile. The time goof could have been avoided if it were clear that you were talking about seconds and your answer was talking about meters/second. I do it too, and got confused too because I wasn't watching the units.

Also, I would suggest breaking the problems up into pieces and solving each piece first rather than getting up all of your substitutions into one big thing. It may make it easier to find any mistakes if you lay it out like this as well as easier to go through your work.

 

[PeroK]

ugh, I am a moron too. btw, can you guys tell me what difficulty you would think this problem is? i would never have figured out the velocities equal each other if no one pointed it out... ugh. i feel like this is too hard a course for me.

I think you chose a complicated way to do it. If you begin at t = 0 when the train starts to move, you get:

$x_t = \frac{1}{2}at^2$

$x_p = v(t-6)$

This is much simpler than your equations.

With this approach, you are looking for v where you can solving the quadratic equation in t (by setting $x_t = x_p$). And, to get the minimum v, you need the quadratic to have exactly one solution:

No solutions => v is too low and she never catches the train
Two solutions => v is too high and she catches up with the train, overtakes it, but it keeps speeding up and then catches her (so there are two times where the distances are equal)
One solution => v is just right and she just catches the train (just as the train reaches her speed)

There are, of course, other ways to solve the problem. Thinking about the equal velocities is a different solution. I wouldn't worry about that for now. I would concentrate on solving the quadratic equation implied above.

The key question is: when does a quadratic equation have exactly one solution?