What Is the Minimum Deceleration to Prevent Two Trains from Colliding?

[showzen]

Homework Statement

Two trains, one traveling 60.0mph east and the other traveling 90.0mph west, are on the same track. When they are 10000 ft apart both train's brakes are applied. If they both have the same deceleration, what is the minimum deceleration to prevent a collision?

Homework Equations

x = x_o + v_ot + 0.5at²

The Attempt at a Solution

I converted velocities to ft/s, 60mph = 88ft/s , 90mph = 132ft/s

I chose the location of the eastbound train to be the origin of the x-axis, then wrote out the positions of both trains as:
x₁ = 88t - 0.5at²
x₂ = 10000 - 132t + 0.5at²

I supposed that the minimum acceleration to prevent a collision implies that x₁ approaches x₂. So
x₁ = x₂
88t - 0.5at² = 10000 - 132t + 0.5at²

Solving for a gives:
a = (88+132)/t - 10000/t²

I am currently stuck here, and really puzzled as to why a is dependent on t. Acceleration must be a constant, so I think I must have some error somewhere that I cannot find. Any help is greatly appreciated.

 

[Hardik Batra]

Welcome to PF.

Use the another equation of kinematics of motion without time.

And Use the relative velocity concept. Here two trains going in opposite direction. What is the velocity of first train relative to second.?

 

[Nathanael]

I am currently stuck here, and really puzzled as to why a is dependent on t.

a is dependent on t simply because that's how you chose to set up the problem. That is, you chose to use time as an intermidiate step ("the middle man")

You have one more constraint (equation) that you're overlooking.

You've set $x_1=x_2$ because you want to find the minimum
(and the minimum is where they just barely don't crash, meaning, $x_1\approx x_2$)

But $x_1=x_2$ is true even if they do crash, isn't it? So what else must be true?

 

[Orodruin]

Hint: If they have the same deceleration, they will not stop at the same time.

 

[showzen]

Okay, I think that I have got the solution.

Using x = x_o + (v² - v_o²) / (2a) :

x₁ = -(88)² / (-2a)
x₂ = 10000 - (132)² / (2a)

x₁ ≈ x₂

I was then able to solve for a and obtain a = 1.26 FT/s².

I see now that I failed to notice some crucial bits of information during my first attempts at this problem.

Thanks everyone for the help!