# Simple kinematics problem

1. Feb 5, 2010

### zero_infinity

1. The problem statement, all variables and given/known data
A train pulls away from a station with constant acceleration of .4 m/s^2. A passenger arrives at a point next to 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 still catch the train?

2. Relevant equations
$$\Delta$$x = v0t + a/2*t2
V2f=V2i+ 2a*$$\Delta$$x

3. The attempt at a solution
I can find the position functions of both objects, but I can't rearrange the formulas just right where I'll end up with just one unknown.

i.e.

Xpassenger = v0t
Xtrain = 7.2 + 2.4*t + .2*t2

I know their positions have to be the same and I know their velocities have to be the same at that point (I picture it as a line touching the tip of a quadratic, like a derivative).

Since it's asking for a minimum, I'm also unsure of how to apply calculus. i.e. What am I taking the minimum of?

Last edited: Feb 5, 2010
2. Feb 5, 2010

### Spinnor

Also v(t) = v_o + a*t , v_o =0 so v(t=6) = ?

3. Feb 5, 2010

### zero_infinity

Um, what?

The passenger is going to be moving with constant speed.

I know the answer - it's 4.8 m/s (i don't know how to arrive at it). That equation doesn't help me for the passenger, and I've calculated the velocity for the train (at t=6).

4. Feb 6, 2010

### zero_infinity

Still need help with this. :/

Nobody knows how to do it? Or did I not follow the rules? I'm pretty sure I did..

5. Feb 6, 2010

### ehild

The positions should be the same, the velocities do not.

Xpassanger=xtrain

This is an equation with two unknown, v0 and t (the time needed for the passenger to reach the train).You need the minimum v0.
It is possible to express v0 as function of t and find t where v0(t) is minimum.

ehild

6. Feb 6, 2010

### zero_infinity

Yes, I have done that too. It's wrong

v(t) = .2(t^2 + 12*t + 36)/t

v'(t) = .2(t^2-36)/t^2 =0

this gives two extrema: 6 and -6

i already know the answer is 4.8, so... it doesn't work like that. I'm also pretty sure the velocities have to be the same.

http://img246.imageshack.us/img246/6751/posy.jpg [Broken]

Last edited by a moderator: May 4, 2017
7. Feb 6, 2010

### ehild

The time must be positive. Just substitute t=6 in v(t) = .2(t^2 + 12*t + 36)/t = .2(6^2 + 12*6 + 36)/6=? I got 4.8, but I might be wrong...

ehild

8. Feb 6, 2010

### zero_infinity

How did you know they were going to meet at t=6(or t=12?)? It doesn't make any sense to me.

That's really not intuitive for me, wtf. If the passenger starts running at t=6, how they possibly meet at t=6(t=12?)?

Last edited: Feb 6, 2010
9. Feb 6, 2010

### ehild

"Xpassenger = v0t
Xtrain = 7.2 + 2.4*t + .2*t2"

Was it you who set up these equations? What does t mean?

ehild

10. Feb 6, 2010

### zero_infinity

.... time

the function for the train could have just been .2t^2(1/2*a*t^2), but i rewrote it for t=6 (found its position, velocity)

if you're plugging 6 into t, then it's 6 extra seconds on-top of the 6 after it started accelerating. how did you know to plug in 6?

11. Feb 6, 2010

### ehild

Yes t is time but the time of what?

ehild

12. Feb 6, 2010

### zero_infinity

what do you mean?!

in that equation it's the time it takes for them to meet at the same point

!?

13. Feb 6, 2010

### ehild

You have set up two equations for the position of the passenger and for the position of the train

"Xpassenger = v0t

Xtrain = 7.2 + 2.4*t + .2*t2"

and these are the same when they meet at tm (meeting time):

v0*tm=7.2 + 2.4*tm + .2*tm^2 ---> v0=7.2/tm + 2.4+ .2*tm.

The minimum value of v0 is needed. You differentiated v0 with respect to tm and have found that it is zero for tm =6 s, so the passenger needs the lowest speed if he caches the train in 6 s.

If you plug in 6 for tm in the expession for v0, you get 4.8 m/s.

v0=7.2/6+ 2.4+ .2*6=1.2+2.4+1.2=4.8.

ehild

14. Feb 6, 2010

### zero_infinity

i'm an idiot

-_-

15. Feb 6, 2010

### ehild

Well.... :rofl: