# Motion of two trains physics problem

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.

rl.bhat
Homework Helper
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?

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?

rl.bhat
Homework Helper
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.

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?

rl.bhat
Homework Helper
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.

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?

Also plugging that number back in I get the acceleration for both of the equations to be

a= -0.5 m/s^2 minimum

rl.bhat
Homework Helper
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.