Train Kinematics Problem

In summary, the minimum deceleration to prevent a collision is when x_1 approaches x_2. x_1 is dependent on t, but x_2 is not. If the two trains have the same deceleration, they will not stop at the same time.
  • #1
showzen
34
0

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 = xo + vot + 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.
 
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  • #2
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.?
 
  • #3
showzen said:
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 [itex]x_1=x_2[/itex] because you want to find the minimum
(and the minimum is where they just barely don't crash, meaning, [itex]x_1\approx x_2[/itex])

But [itex]x_1=x_2[/itex] is true even if they do crash, isn't it? So what else must be true?
 
  • #4
Hint: If they have the same deceleration, they will not stop at the same time.
 
  • #5
Okay, I think that I have got the solution.

Using x = xo + (v2 - vo2) / (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!
 

1. What is train kinematics?

Train kinematics is the study of the motion of trains, including their speed, acceleration, and position over time.

2. How is train kinematics used in real life?

Train kinematics is used in the design and operation of trains to ensure safe and efficient transportation. It is also used in train scheduling and route planning.

3. What are the key equations used in train kinematics?

The key equations used in train kinematics are the equations of motion, including velocity = distance/time, acceleration = change in velocity/time, and position = initial position + velocity*time + (1/2)*acceleration*time^2.

4. What factors affect train kinematics?

The factors that affect train kinematics include the train's weight, engine power, track conditions, and external forces such as wind resistance.

5. How does train kinematics differ from other types of kinematics?

Train kinematics differs from other types of kinematics in that trains typically operate on a fixed track and have limited motion in one dimension, whereas other types of kinematics may involve more complex movements in multiple dimensions.

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