Motion of two trains physics problem

  • #1
Two trains heading straight for each other on the same track are 850m apart when their engineers see each other and hit the brakes, giving both trains a constant deceleration (a). The Express is heading west at a speed of 15.0 m/s, while the east bound Flyer is traveling at a speed of 25 m/s. Calculate the minimum deceleration (a) required for a collision to be avoided.

V1 speed of first train = 15.0 m/s
V2 speed of second train = 25 m/s
x=850 m

Can someone please help me solve this problem. I have been staring at it for about an hour and cannot even begin to go anywhere.
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
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What is the relevant kinematic equation which relates vi, vf, a and x?
In this problem what is the final velocities of the trains?
If west bound train moves a distance x m before it stops, what is the distance traveled by east bound train?
 
  • #3
What is the relevant kinematic equation which relates vi, vf, a and x?
In this problem what is the final velocities of the trains?
If west bound train moves a distance x m before it stops, what is the distance traveled by east bound train?
The kinematic equation would be..

V^2 = V0^2 + 2a (X-X0)

The final velocity would have to be 0 since they have to come to a complete stop to avoid a collision.

I am not really sure about the distance.. since the total distance is 850..possibly 850-x?
 
  • #4
rl.bhat
Homework Helper
4,433
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The kinematic equation would be..

V^2 = V0^2 + 2a (X-X0)

The final velocity would have to be 0 since they have to come to a complete stop to avoid a collision.

I am not really sure about the distance.. since the total distance is 850..possibly 850-x?
Yes. You are right. Now wright down two equations for two trains. And solve for x. From that you can find a.
 
  • #5
Yes. You are right. Now wright down two equations for two trains. And solve for x. From that you can find a.
ok let me try this

15^2 = 2a(850-x)

and

25^2 = 2a(850-x)

a = 25^2 / (1700 - 2x) plug back into eqn. 1? to solve for x?
 
  • #6
rl.bhat
Homework Helper
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ok let me try this

15^2 = 2a(850-x)

and

25^2 = 2a(850-x)

a = 25^2 / (1700 - 2x) plug back into eqn. 1? to solve for x?
First one should be
15^2 = 2ax.
Now 15^2/25^2 = 2ax/2a(850-x)
Solve for x.
 
  • #7
First one should be
15^2 = 2ax.
Now 15^2/25^2 = 2ax/2a(850-x)
Solve for x.
ok I get x = 225m is this correct? It was a little difficult to solve for x. The acceleration should cancel correct?
 
  • #8
Also plugging that number back in I get the acceleration for both of the equations to be

a= -0.5 m/s^2 minimum
 
  • #9
rl.bhat
Homework Helper
4,433
7
Also plugging that number back in I get the acceleration for both of the equations to be

a= -0.5 m/s^2 minimum
Yes. You are right.
 

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