Solving Train Collision: Min Rate, Collision Speed & Distance Traveled

[ts21121]

A passenger train is traveling at 30 m/s when the engineer sees a freight train 336 m ahead of his train traveling in the same direction on the same track. The freight train is moving at a speed of 5.9 m/s.

1.If the reaction time of the engineer is 0.36 s, what is the minimum (constant) rate at which the passenger train must lose speed if a collision is to be avoided?

2. If the engineer's reaction time is 0.79 s and the train loses speed at the minimum rate described in Part (a), at what rate is the passenger train approaching the freight train when the two collide?


For both reaction times, how far will the passenger train have traveled in the time between the sighting of the freight train and the collision?

x=x₀+v₀t +1/2 at²
v=v₀+at

The Attempt at a Solution

for part 1 i know the answer is .887m/s^2.
after this i am completely lost on how to solve for part 2 and 3

 

[rl.bhat]

Hi ts21121, welcome to PF.
I am getting slightly different answer for the part 1. Will you show your calculations?

 

[kerimek]

.

I would approach this problem by using the principles of motion and kinematics to solve for the minimum rate at which the passenger train must lose speed and the distance traveled before collision.

1. To solve for the minimum rate at which the passenger train must lose speed, we can use the formula v = v0 + at, where v is the final velocity, v0 is the initial velocity, a is the acceleration, and t is the time. In this case, we know the initial velocity (30 m/s), the final velocity (5.9 m/s), and the time (0.36 s). We can rearrange the formula to solve for the acceleration (a), which will give us the minimum rate at which the passenger train must lose speed.

a = (v - v0)/t = (5.9 m/s - 30 m/s)/0.36 s = -83.33 m/s^2

Since acceleration is a vector quantity, the negative sign indicates that the train must decelerate in order to avoid a collision. Therefore, the minimum rate at which the passenger train must lose speed is 83.33 m/s^2.

2. To solve for the rate at which the passenger train is approaching the freight train when the two collide, we can use the formula x = x0 + v0t + 1/2at^2, where x is the distance traveled, x0 is the initial position, v0 is the initial velocity, a is the acceleration, and t is the time. In this case, we know the initial position (336 m), the initial velocity (30 m/s), the acceleration (-83.33 m/s^2), and the time (0.79 s). We can rearrange the formula to solve for the distance traveled (x), which will give us the distance between the two trains when they collide.

x = x0 + v0t + 1/2at^2 = 336 m + (30 m/s)(0.79 s) + 1/2(-83.33 m/s^2)(0.79 s)^2 = 5.9 m

Therefore, the two trains will collide when the passenger train has traveled 5.9 m.

3. To solve for the distance traveled by the passenger train between the sighting of the freight train and the collision, we can use the same formula as in