# What Deceleration Prevents a Train Collision?

• soul5
In summary: is coming closer and closer to your train, then you would need to give it more and more space to avoid a collision.
soul5

## Homework Statement

When a high-speed passenger train traveling at 158 km/h rounds a bend, the engineer is shocked to see that a locomotive has improperly entered onto the track from a siding and is a distance D = 699 m ahead. The locomotive is moving at 27.2 km/h. The engineer of the high-speed train immediately applies the brakes. What must be the magnitude of the resulting constant deceleration if a collision is to be just avoided?

## Homework Equations

x = xi + vi * t + 1/2 * a * t ^2

## The Attempt at a Solution

I think you can set that equation equal to each other I don't understand the black part please help me. I think I can use that equation, but how do I find the time? How would I do this question?

Is it just me or is this question some how flawed?

well first off we need to know the direction of the locomotive that is traveling at 27.2 km/h before we can do too much.

but basically you are looking for a=? . what must the deceleration of the train be to stop before it hits the locomotive?

This is a somewhat poorly worded question. You have to assume that (a) the locomotive continue moving at 27.2 km/hr and (b) that the locomotive is moving in the same direction as the high-speed train. You don't need the train to come to a stop. You merely need to make it slow down to 27.2 km/hr at exactly the point where the distance between the train and locomotive is zero.

D H said:
This is a somewhat poorly worded question. You have to assume that (a) the locomotive continue moving at 27.2 km/hr and (b) that the locomotive is moving in the same direction as the high-speed train. You don't need the train to come to a stop. You merely need to make it slow down to 27.2 km/hr at exactly the point where the distance between the train and locomotive is zero.

how would I do that if I don't have the time?

You have two unknowns: The acceleration a and the time of closest approach tc. That means you need two independent equations. You know the desired speed and separation distance at this time. So, write equations that describe the speed and separation as a function of acceleration and time. That's equations, from which you can derive the requisite acceleration (and time as a by-product).

if I were you I wouldn't start from here …

Hi soul5!
soul5 said:
x = xi + vi * t + 1/2 * a * t ^2 .

I think I can use that equation, but how do I find the time?

You have the initial and final speeds ui and uf, and the distance, s.

You want the acceleration, a, and you're not asked to find the time.

What equations do you know that involve only ui uf a and s?

That won't work in this case, Tim, because you don't know the distance s.

D H said:
That won't work in this case, Tim, because you don't know the distance s.

Hi D H!

There is a frame in which s = 699

That's the wrong distance. Suppose the engineer followed your advice, breaking such that the train achieves a speed of 27.2 km/h after traveling 699 meters. Meanwhile, the locomotive will have moved further down the track. The goal is to just avoid a collision. Your approach is overkill.

Depends on the frame.

Everything is relative.

I think I know now v^2 = vi ^2 * 2a (x-xo) do u use this formula?

No. That's not even a valid formula. The left hand side has units of length2/time2 while the right hand side has units of length4/time4.

soul5 said:
I think I know now v^2 = vi ^2 * 2a (x-xo) do u use this formula?

We've learn like only 3 formulas if this isn't it I don't know what it is. Can someone tell me what forumla to use?

What are the other two? (You posted one for distance in the original post.) Do you know one for velocity?

With v^2 = vi ^2 * 2a (x-xo) you must be referring to:

$$(v_{f})^2 - (v_{i})^2 = 2a(x - x_{0})$$

You have the initial velocity (158 km/h), the final velocity you need (27.2 km/h) and the distance between the two objects.

Assuming the locomotive and the train are moving in the same direction, your acceleration (or deceleration, if you prefer) looks like:

[tex]
a = \frac{(v_{f})^2 - (v_{i})^2}{2(x - x_{0})}
[/itex]

Note: the negative sign of the acceleration means you are slowing down.

Note 2: If the locomotive is coming at your train... then that's a whole new problem. i.e. you must define the meaning of "avoding collision" first. Then you must consider how much of "safe distance" you want to give to your train - if the locomotive is moving at 27.2 km/h, giving your train a second or two wouldn't make much sense, right? Thus, safe distance of 38m or less are out of the question (as the locomotive is moving in at 7.56 m/s) Say you give yourself a good 100 meters of safe distance. That means the distance your train covers + the distance the locomotive covers is less than or equal to (in maximum case) 599 meters, and $t_{c}$ is same for both. Also, what is the final velocity you want to reach at? Ideally, you'd want to stop, but you need to know how much of a distance you have as a cushion, so you may not use the equation above.

Considering the complexity this will bring, I have hard time convincing myself that the locomotive is actually coming at your train. But who knows? Maybe food for thought.

Last edited:
soul5 said:
I think I know now v^2 = vi ^2 * 2a (x-xo) do u use this formula?

Hi soul5!

As brasidas says, this should be v^2 = vi ^2 + 2a (x-xo).

But you can't use it with x-x0 = 699, because as D H says, that will stop your train when it reaches where the other train was
D H said:
That's the wrong distance. Suppose the engineer followed your advice, breaking such that the train achieves a speed of 27.2 km/h after traveling 699 meters. Meanwhile, the locomotive will have moved further down the track. The goal is to just avoid a collision. Your approach is overkill.

x and x0 in this formula have to be measured relative to a stationary point.

You need to use relative speeds … in other words, pretend that the whole track is moving backwards at 27.2 km/h, which won't change the acceleration, but will make the other train stationary.

(This is Newton's Principle of Relativity, which says that the same equations of motion are valid in any uniform-velocity frame)

## 1. What is constant acceleration of a train?

Constant acceleration of a train refers to the unchanging rate at which the train's velocity increases or decreases over time. It is measured in meters per second squared (m/s²).

## 2. How is constant acceleration of a train calculated?

The constant acceleration of a train can be calculated using the formula a = (vf - vi)/t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

## 3. What factors affect the constant acceleration of a train?

The constant acceleration of a train can be affected by several factors, including the power and design of the train's engine, the weight and mass of the train, and external forces such as friction and air resistance.

## 4. Why is constant acceleration important for trains?

Constant acceleration is important for trains because it allows them to reach and maintain a desired speed, as well as to slow down or stop efficiently and safely. It also helps to reduce travel time and improve overall efficiency.

## 5. How does constant acceleration affect the passengers on a train?

Constant acceleration can have an impact on passengers by causing them to experience a feeling of being pushed back or pulled forward, depending on the direction of acceleration. It can also affect the comfort level of the ride, as sudden changes in acceleration can result in jerky movements.

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