Subway Kinematics motion

1. Dec 15, 2013

negation

1. The problem statement, all variables and given/known data

You're an investigator for the National Transportation safety board, examining a subway accident in which a train going at 80kmh^-1 collided with a slower train traveling in the same direction at 25kmh^-1. Your job is to determine the relative speed of the collision to help establish new crash standards. The faster train's black box shows that it began negatively accelerating at 2.1ms^-1 when it was 50m from the slower train, while the slower train continued at constant speed.
What do you report?

2. Relevant equations
None.

3. The attempt at a solution

Retrying attempt

Last edited: Dec 15, 2013
2. Dec 15, 2013

Curious3141

The problem with this calculation is that when you convert 2.1 m/s^2 into km/h^2, it's actually:

$\frac{2.1}{1000} * 3600^2 = 27216 kmh^{-2}$

remember the square on the time unit denominator. You should get a more plausible answer this way.

Of course, it's easier if you convert the speed into $ms^{-1}$ rather than attempting this.

In any case, is it really necessary to solve for the stopping distance? Why not try to calculate the time the two trains occupy the same location (set up a common coordinate axis for this), and then solve for the speed of the first train at this time of collision?

3. Dec 15, 2013

Staff: Mentor

Try this: Express the position of each train as a function of time, starting from the moment that the slower train is 50 m ahead. Then you can solve for the time when they collide.

4. Dec 15, 2013

negation

dx = vit + 0.5at^2

train 1: dx1 = 22.2ms^-1 t + 0.5(-2.1ms^-2)t^2
dx1 = 22.2 ms^-1 t - 1.05ms^-2 t^2

train 2: dx2 =50m + 6.9ms^-1 t

d1 = d2

22.2ms^-1 t - 1.05ms^-2 t^2 = 50m + 6.9ms^-1 t

t = 9s

Last edited: Dec 15, 2013
5. Dec 15, 2013

negation

t = 9s.

6. Dec 15, 2013

negation

edit

Last edited: Dec 15, 2013
7. Dec 15, 2013

Staff: Mentor

OK.

Your equation is OK, but not your solution. Try one more time.

8. Dec 15, 2013

negation

I'm getting t = 9.6s and t = 4.9s.
The quadratic equation 1.05ms^-2 (t^2) - 15.3ms^-1 (t) + 50m has 2 real roots.

9. Dec 15, 2013

Staff: Mentor

Good. The shorter time is the one you want. Find the speed of the fast train at that time and then the relative speed.

10. Dec 15, 2013

negation

Could you expound on what is meant conceptually when you state "Find the speed of the fast train at that time and then the relative speed."?
Speed of the train implies vi and the relative speed implies vf?
Also, why do we want 4.9s and not 9.6s?

11. Dec 15, 2013

negation

dx = vit + 0.5at^2

50m = 4.6t - 1.05ms^-2(4.6)^2
vi = 15.7ms^-1 (is vi what you referred to as speed of the train at t = 4.6s?)

And if relative speed implies vf, then;

dx = 0.5(vi + vf)t
50m = 0.5(15.7ms^-1 + vf)4.6
vf = 21.6kmh^-1

12. Dec 15, 2013

Staff: Mentor

You want the speed of the trains when they collide, so you can calculate the relative speed. You know the initial speed of the fast train and its acceleration, so find the speed at the time of collision.

That's when they first collide. After that point, the kinematic equations no longer apply.

13. Dec 15, 2013

negation

vf^2 - vi^2 = 2a(dx)
vf^2 - (22.2)^2 = 2(-2.1)(50)
vf = 16.8ms^-1 = 61kmh^-1

Sounds logical

Last edited: Dec 15, 2013
14. Dec 15, 2013

Staff: Mentor

50m is the initial separation, not the distance traveled. Use the time.

15. Dec 15, 2013

negation

Edit

Last edited: Dec 15, 2013
16. Dec 15, 2013

negation

Solved

at t = 4.6s:

dx = vit + 0.5at^2
dx = 80m

Where train 1 is moving at 80kmh^-1 and train 2 moving at 25kmh^-1, relative to train 1, train 2 is moving at 0kmh^-1. Relative to train 2, train 1 moves at (80kmh^-1 - 25kmh^-1) = 55kmh^-1

vf^2 - vi^2 = 2a.dx
vf^2 - (15.3ms^-1)^2 = 2(-2.1ms^-2)(80m)
vf = 36kmh^-1

17. Dec 15, 2013

Staff: Mentor

Answer this: At t = 0, the speed is 22.22 m/s. You are given the acceleration. What is the speed after t = 4.97 seconds? (Note the more accurate time.)

18. Dec 15, 2013

negation

Speed decreases as time, t, tends towards infinity.

How did you get 4.97?
Solving via quadratic I'm only about to get 4.95 rounded off to 3sf.

Last edited: Dec 15, 2013
19. Dec 15, 2013

Curious3141

I'm getting 4.97s as well. You shouldn't round off intermediate results, or you should round to at least 2sf more than you need in your final answer. With a scientific calc, you don't even need to round off (more than the machine does internally), because the memory can store the results of your calculation quite precisely.

Speed doesn't decrease indefinitely, since train 1 will eventually come to a stop if train 2 wasn't there. As it happens, it never gets a chance to stop because it hits train 2 at a certain speed. Your task is to find that speed.

You already have the time of collision t. You know the initial speed of train 1 (80/3.6 m/s -> notice that I'm leaving numbers in exact form wherever possible). You know the deceleration of train 1 (a = -2.1m/s). Haven't you learnt a simple kinematic equation relating initial speed, final speed, acceleration and time? Apply that.

20. Dec 15, 2013

negation

It's just vf = vi + at.
vf = 22.2ms^-1 + (-2.1ms^-2)(4.97s)

Would it be possible to solve this problem using local minimization? The many steps are laborious and inelegant.