- #1

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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.

- Thread starter ur5pointos2sl
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- #1

- 96

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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.

- #2

rl.bhat

Homework Helper

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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

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The kinematic equation would be..

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?

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

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Yes. You are right. Now wright down two equations for two trains. And solve for x. From that you can find a.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?

- #5

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ok let me try thisYes. You are right. Now wright down two equations for two trains. And solve for x. From that you can find a.

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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First one should beok 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?

15^2 = 2ax.

Now 15^2/25^2 = 2ax/2a(850-x)

Solve for x.

- #7

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ok I get x = 225m is this correct? It was a little difficult to solve for x. The acceleration should cancel correct?First one should be

15^2 = 2ax.

Now 15^2/25^2 = 2ax/2a(850-x)

Solve for x.

- #8

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a= -0.5 m/s^2 minimum

- #9

rl.bhat

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Yes. You are right.

a= -0.5 m/s^2 minimum

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