What is the acceleration of the train.

[astrololo]

Homework Statement

A train is going towards a stationary car at a speed of 50 km/h. There is a distance of 250 m between the train and the car. What must be the acceleration (which is constant) of the train so that the train won't collide with the car ?

Initial position of train = 0

final position = 0,250 km - x

Speed= 50 km/h

Distance between car and train = 0,250 km[/B]

Homework Equations

Vfinal ^2 = v initial ^ 2 + 2*acceleration*(x final - x initial)

The Attempt at a Solution

0 = 50^2 + 2 * a * ((0,250-x)-0)

0=2500 + 2a(0,250-x)

I'm kinda stuck :/[/B]

 

[Bystander]

What's "v_final?"

 

[astrololo]

What's "v_final?"

It's your final speed.

 

[Bystander]

"You" are the train. What is "v_final?"

 

[astrololo]

"You" are the train. What is "v_final?"

Well it's 0 km/h because I need to stop before 250 m or else I'm going to destory the car.

 

[Bystander]

You're given velocity in km/h and distance in meters. Do you need to do any conversions?

 

[astrololo]

You're given velocity in km/h and distance in meters. Do you need to do any conversions?

Yes, I converted my 250 m to km which is 250 m * 1 km / 1000 m =0,250 km

 

[Bystander]

Okay, European convention. What's "x?"

 

[astrololo]

Okay, European convention. What's "x?"

It's the position where I'll stop the train before it touches the car.

 

[Bystander]

You've a quarter kilometer range. Does the problem statement ask for the general solution? Or, a specific value?

 

[astrololo]

You've a quarter kilometer range. Does the problem statement ask for the general solution? Or, a specific value?

It's asking for a specific value. I could take any value of x where the train would stop before reaching the car, but I think we're really asking for a specific value where the train ALMOST reaches the car. So we're really asking for the "maximum" value of x I think.

 

[billy_joule]

0=2500 + 2a(0,250-x)

What is x here?
x _final - x _initial is given as 250 m in the problem statement so there's no need to have that x there.
So you have one equation with one unknown, you can solve for a. Make sure you include the correct units for your answer.

EDIT: Oh, I see they want the answer in terms of x.

 

[Bystander]

So we're really asking for the "maximum" value of x I think.

So do I. You've picked an "awkward" unit for "a" as your work is currently set up. Might be best to consider billy joule's recommendation.

 

[astrololo]

So do I. You've picked an "awkward" unit for "a" as your work is currently set up. Might be best to consider billy joule's recommendation.

Ok, so I get :

0=2500 + 2ax

-2500=2ax

-1250/x=a

 

[billy_joule]

That's not consistent with the use of x in the OP:

final position = 0,250 km - x

In words, x is the distance from the car to the where the train stops, so the maximum value of x is 250m, which corresponds to an infinite acceleration ie the train stops instantaneously. (x _{final position} = 0.250 km - x = 0) Not likely the answer they're looking for.

For these sorts of problems it's usually safe to assume they want the answer when x = 0. That is, the train comes to a stop 'touching' the car. That corresponds to the minimum acceleration to avoid a crash.

 

[billy_joule]

Note that x is used three times with three different meanings:

x _{final position} = 250m - x'
x_{initial position} = 0

Where x' is the final train to car distance

 

[astrololo]

Note that x is used three times with three different meanings:

x _{final position} = 250m - x'
x_{initial position} = 0

Where x' is the final train to car distance

So I was right in writing it this way ?

 

[billy_joule]

Which way?

Where exactly does the problem statement end and your work begin?

Is this part of the problem statement or your own work?

Initial position of train = 0

final position = 0,250 km - x

If that's your own work, there was no need to introduce the additional variable x' that describes the final train-car distance. Just solve assuming the train comes to a stop at the car.
If it is part of the problem statement and they want an answer in terms of x' then your attempt in the OP is very close, just rearrange for a.

 

[astrololo]

Which way?

Where exactly does the problem statement end and your work begin?

Is this part of the problem statement or your own work?If that's your own work, there was no need to introduce the additional variable x' that describes the final train-car distance. Just solve assuming the train comes to a stop at the car.
If it is part of the problem statement and they want an answer in terms of x' then your attempt in the OP is very close, just rearrange for a.

no, that's my attempt at the problem. The problem doesn't provide this. I did it.

 

[astrololo]

Which way?

Where exactly does the problem statement end and your work begin?

Is this part of the problem statement or your own work?If that's your own work, there was no need to introduce the additional variable x' that describes the final train-car distance. Just solve assuming the train comes to a stop at the car.
If it is part of the problem statement and they want an answer in terms of x' then your attempt in the OP is very close, just rearrange for a.

You mean this : 0=2500+2a 0,250 ? I obtain -5000 of acceleration

 

[billy_joule]

but what are the units?

 

[astrololo]

but what are the units?

0 km/h = 50 km/ h ^2 + 2 a (0,250 km - 0 km)

 

[billy_joule]

Correct. So what are the units of your answer?

 

[astrololo]

Correct. So what are the units of your answer?

wait are u telling me that my acceleration units are km/h^2 ?

 

[astrololo]

Correct. So what are the units of your answer?

Ok yeah, it's working now ! I got the answer ! The problem really was asking to find the acceleration at moment the train was going to percute the car. In other words, they wanted the maximum value for x and not any x before 250 m.

 

[billy_joule]

Great, Good work.