How Fast Were the Trains Moving at the Point of Collision?

[negation]

Homework Statement

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?

Homework Equations

None.

The Attempt at a Solution

Retrying attempt

 

[Curious3141]

Homework Statement

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?

Homework Equations

None.

The Attempt at a Solution

Determine the braking distance required of train traveling at 80kmh^-1:

vf^2 - vi^2 = 2(-7.56kmh^-2)(yf - yi)
yf - yi = dy = 423km

Something is wrong: 423 km?

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?

 

[Doc Al]

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.

 

[negation]

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?

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

 

[negation]

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.

t = 9s.

 

[negation]

edit

 

[Doc Al]

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

OK.

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


d1 = d2

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

t = 9s

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

 

[negation]

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

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.

 

[Doc Al]

I'm getting t = 9.6s and t = 4.9s

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

 

[negation]

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

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?

 

[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

 

[Doc Al]

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?

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.

Also, why do we want 4.9s and not 9.6s?

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

 

[negation]

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.

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

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

Sounds logical

 

[Doc Al]

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

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

 

[negation]

Edit

 

[negation]

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

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

 

[Doc Al]

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.)

 

[negation]

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.)

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.

 

[Curious3141]

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.

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 learned a simple kinematic equation relating initial speed, final speed, acceleration and time? Apply that.

 

[negation]

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 learned a simple kinematic equation relating initial speed, final speed, acceleration and time? Apply that.

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.

 

[Curious3141]

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.

You still haven't stated a final answer. Remember what I said about round-off errors, you should try to work to greater precision in all your intermediate steps.

And I think it would be good to give your final answer in km/h, which is the speed unit the question was posed in. Remember you're pretending to model a "real life" situation, where km/h (or mph in some places) is the more common unit.

As for an alternative solution, I would suggest the problem not be overcomplicated. These steps are the simplest, most direct way to solve it.

 

[negation]

You still haven't stated a final answer. Remember what I said about round-off errors, you should try to work to greater precision in all your intermediate steps.

And I think it would be good to give your final answer in km/h, which is the speed unit the question was posed in. Remember you're pretending to model a "real life" situation, where km/h (or mph in some places) is the more common unit.

As for an alternative solution, I would suggest the problem not be overcomplicated. These steps are the simplest, most direct way to solve it.

Got it!
I realized my units and conversion were the culprits.