Do the trains collide or stop before collision?

  • Thread starter Yapper
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The speeds at which they collide are then 30 m/s for the red train and 10 m/s for the green train. But that's not what the question asks for, so we have to subtract their braking distances from their positions at the time of collision. v0=0; v=at => t=70/20=3.5 s (red train)
  • #1
Yapper
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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
 
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  • #2
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
 
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  • #3
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.
 
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  • #4
umm.. I think that first approach is not so correct, because the collision occurs before the red train stops...
relatve a is 1-(-1)
 
  • #5
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
 
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  • #6
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.
 
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  • #7
K, thanks for all the help. Ill make sure to point out the problem to my prof.
 
  • #8
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! ^^
 
  • #9
new equation:
d(t) = 870-60t+t^2; t<=20 s
d(t) = 70-20(t-20)+0.5(t-20)^2 t>20
 

1. What is the "2 trains headed for eachother" problem?

The "2 trains headed for eachother" problem is a common physics problem used to demonstrate concepts of relative motion and distance. It involves two trains moving towards each other on the same track at different speeds, and asks for the time and distance it takes for them to meet.

2. How do you solve the "2 trains headed for eachother" problem?

To solve the "2 trains headed for eachother" problem, you can use the formula d = rt, where d is distance, r is rate (or speed), and t is time. You can set up two equations for each train and then use algebra to solve for the time it takes for them to meet.

3. What are some real-life applications of the "2 trains headed for eachother" problem?

The "2 trains headed for eachother" problem has many practical applications, such as calculating the time and distance for two cars traveling towards each other on the same road, or for two boats approaching each other in opposite directions on a river. It can also be used to determine the time and distance for two people walking towards each other on a sidewalk.

4. What assumptions are made when solving the "2 trains headed for eachother" problem?

When solving the "2 trains headed for eachother" problem, it is assumed that the trains are moving on the same track, that they start at the same time, and that there are no external factors (such as stops or delays) that would affect their speeds or distances.

5. How can the "2 trains headed for eachother" problem be made more complex?

The "2 trains headed for eachother" problem can be made more complex by adding additional variables, such as one train starting at a different time or location, or by changing the rates of the trains at different points in their journey. It can also be made more complex by adding multiple trains or objects moving towards each other on the same track.

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