Do the trains collide or stop before collision?

  • Thread starter Thread starter Yapper
  • Start date Start date
  • Tags Tags
    Trains
Click For Summary

Homework Help Overview

The problem involves two trains, a red train traveling at 72 km/h and a green train at 144 km/h, approaching each other on a straight track. They begin braking when they are 870 m apart, with a deceleration of 1.0 m/s². The question is whether they collide and, if so, to determine their speeds and separation at the point of collision.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the initial velocities and braking distances of both trains, with some attempting to calculate stopping distances and times to assess potential collision.
  • There is a focus on relative motion and distance equations, with attempts to derive when the distance between the trains becomes zero.
  • Some participants express confusion regarding the timing of the collision relative to the stopping of the red train.
  • Questions arise about the validity of the equations used and the assumptions made regarding the trains' motions.

Discussion Status

The discussion is ongoing, with various interpretations of the problem being explored. Some participants have provided calculations that suggest the red train stops before a collision occurs, while others argue that the collision happens before the red train comes to a complete stop. There is no clear consensus, but several productive lines of reasoning are being examined.

Contextual Notes

Participants note discrepancies in their calculations and the provided answers from the homework platform. There is an acknowledgment of potential errors in the relative distance equations and assumptions about the motion of the trains after braking begins.

Yapper
Messages
101
Reaction score
0
For phys 170 we do assignments on a website from the textbook edugen.wiley.com.

Homework Statement



A red train traveling at 72 km/h and a green train traveling at 144 km/h are headed toward each other along a straight, level track. When they are 870 m apart, each engineer sees the other's train and applies the brakes. The brakes slow each train at the rate of 1.0 m/s^2. Is there a collision?
If so, give (a) the speed of the red train, (b) the speed of the green train, and (c) the separation between the trains when they collide (0 m).
If not, give (a) the speed of the red train (0 m/s), (b) the speed of the green train (0 m/s), and (c) the separation between the trains when they stop.

Homework Equations


v0= initial velocity
v=v0+at
dx=(1/2)at^2 + v0t


The Attempt at a Solution


72km/h = 20 m/s 144 km/h = 40 m/s

Ok so I checked the stopping distances and the red train will stop after 20 seconds, 200 m farther down and after that 20 sec the green train will be (1/2)(-1)20^2 + 40(20) = 600m farther down the track, so I assumed the red train's velocity to be 0m/s since it stops before the collision.

Then for the green train I set 1/2(-1)x^2 +40x = 670, to see how long it would take for the green train to hit the stationary red train, and solved for x. I got 23.8755 seconds. Then solved for velocity 40-23.8755 =16.1245 m/s.
So I inputed a) 0 m/s b) 16.1245 m/s c) 0 m

The website said I was wrong and after many different tries it told me that the answer was:
(a)
Number: 3.127717305696
Units: m/s
(b)
Number: 23.127717305696
Units: m/s
(c)
Number: 0
Units: m

Please help, What did I do wrong? or is the website wrong
 
Physics news on Phys.org
haven't got anywhere..

but I am pretty sure that you need to use relative distance for these like questions.
So, find an equation for the distance between two trains. I got:
d(t) = 870-60*t+t^2 (using relative a and v)
and it says d(t) is min when d(t) = -30, and t = 30,
so that means they collided,

and d(t) = 0 when t = 24.52277442

according to that answer relative v is 20 when d btw them is 0,
but mine equations says relative v is 10.. so try finding some error in my way
 
Last edited:
I did it first that way but then I realized that the red train stops before 24.522, so for those extra 4.552 seconds its moving backwards, which shouldn't be possible solely from the application of brakes. And based on the answer the time to collision is 40-23.127 which is like 16.something seconds which I don't get near at all. And isn't the relative velocity going to be 60-2t(until red train stops) since both trains are decelerating away from each other at 1m/s^s and relative velocity is 60m/s to start with.
 
Last edited:
umm.. I think that first approach is not so correct, because the collision occurs before the red train stops...
relatve a is 1-(-1)
 
according to them yes, but they don't show there work and I have no idea how they figure the trains hit before the red train stops seeing as how after 20 sec the green train has closed 600 m and the red train has closed 200m and stopped, and they started 870m apart...

((-1m/s^2)(20s)^2)/2 +20m/s*20s = 200 m ((-1m/s^2)(20s)^2)/2 +40m/s*20s = 600m 600+200=800 which leaves 70m between them at t = 20 sec
 
Last edited:
Yeah, I think something's wrong with the question... Yapper, I'm getting exactly the same answer as you...

My work:

I'm getting that the red train stops in 20s... which is at a distance of 200m... In that 20s, the green train travels 40(20)+1/2(-1)(20)^2 = 600m... so there's still a 70m gap between them... the collision still takes place, but the red train is stopped... so the velocity of the red train should be 0.

The green train travels a distance of 70m before the collision takes place... using v2^2 = v1^2 + 2as, I get the velocity of the green train as 16.1245 m/s when the collision takes place.
 
Last edited:
K, thanks for all the help. Ill make sure to point out the problem to my prof.
 
I guess mine relative distance equation is wrong because acc. to it, the green train starts traveling backwards after 20 s..
now, i know why it used to take so much time to solve these like problems.

thanks! - that other approach is much better than mine! ^^
 
new equation:
d(t) = 870-60t+t^2; t<=20 s
d(t) = 70-20(t-20)+0.5(t-20)^2 t>20
 

Similar threads

Replies
3
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
20
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 2 ·
Replies
2
Views
7K
  • · Replies 4 ·
Replies
4
Views
2K