Solve for Braking Acceleration for Emergency Stop

In summary: I am taking a break for the night on it. Email if you can explain the hint a little bit more. Will be at work at it again in the morning. I know it isn't that hard of a problem but I am making it hard. I just need to figure out what I am doing wrong with the conversion for it. Thank you for your help.In summary, the conversation revolves around a problem of finding the braking acceleration of a train traveling at 297 km/h and requiring 1.45 km to come to an emergency stop. The conversation includes the use of various formulas and equations, and ultimately concludes with the correct solution of -2.35 m/s^2 for the acceleration.
  • #1
rculley1970
28
0
I am having problems with starting a certain problem.

A train traveling at 297 km/h requires 1.45 km to come to an emergency stop. Find the braking acceleration, assuming constant acceleration.

Now I am not given the acceration or time so this one is stumping me. I have tried several formulas including:

deltaX = 1/2(a)(t)^2 + Vo(t)
(v)^2 = (Vo)^2 + 2(a)(delta X)
v = Vo +a(t)

I cannot figure out how to get time or acceleration to solve for the other. The acceleration isn't due to gravity so it isn't (-9.8m/s^2) so I am at a loss for what equation to use. Should I solve for time first? If so, what is the equation I am missing? As far as I know, I am given Vo (297), Vfinal (0), delta Y (-297), delta X (1.45) and I have already tried converting km/h to m/s which the answer is supposed to be in. CONFUSED!

Please help.
 
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  • #2
rculley1970 said:
I am having problems with starting a certain problem.

A train traveling at 297 km/h requires 1.45 km to come to an emergency stop. Find the braking acceleration, assuming constant acceleration.

Now I am not given the acceration or time so this one is stumping me. I have tried several formulas including:

deltaX = 1/2(a)(t)^2 + Vo(t)
(v)^2 = (Vo)^2 + 2(a)(delta X)
v = Vo +a(t)

I cannot figure out how to get time or acceleration to solve for the other. The acceleration isn't due to gravity so it isn't (-9.8m/s^2) so I am at a loss for what equation to use. Should I solve for time first? If so, what is the equation I am missing? As far as I know, I am given Vo (297), Vfinal (0), delta Y (-297), delta X (1.45) and I have already tried converting km/h to m/s which the answer is supposed to be in. CONFUSED!

Please help.

You wrote the equation that you need!
[itex] v_f^2 = v_i^2 + 2 a_x \Delta x [/tex]! That's all you need!
 
  • #3
I tried that already but will try again. I am using Vf^2 = 0 since that is the final velocity, 297 as the initial velocity, and 1.45 as delta X. I am coming up with -30417 but will keep messing with it to figure it out. I know it CAN'T be this hard.

Should I be finding the time for it to stop?
 
  • #4
all u need is

(v)^2 = (Vo)^2 + 2(a)(delta X)

the point here is, to think what happens when u hit brakes. u slow down, which is a deceleration or a negative acceleration. (opposite is speeding up, which is positive acceleration). usually in a problem like this both these situation are referred by the term acceeleration and leave you to decide.

gravity does not come in. you're moving horizontally. gravity acts only on objects traveling in vertical direction.

so from starting velocity V0=297 km/h to final velocity V=0 (i.e. to a stop)

0 = V0^2 + 2 a (delta X)

u know delta X. plug in and solve for 'a', acceleration.
final answer will be negative, proving that you are actually decelerating.
 
  • #5
OK, I have the equation:

0^2 = (82.5)^2 + 2(a)1450

changed km/h to m/s and km to m.

I am coming up with -2.35m/s acceleration. If I am wrong let me know. I am busy checking it right now.
 
  • #6
rculley1970 said:
OK, I have the equation:

0^2 = (82.5)^2 + 2(a)1450

changed km/h to m/s and km to m.

I am coming up with -2.35m/s acceleration. If I am wrong let me know. I am busy checking it right now.

right...except for the units...
 
  • #7
I know it is probably an easy problem after seeing how it is done but I have been fighting this problem by myself for 4 days and just can't seem to figure out why I can't get the acceleration without the the time. The example in the book shows:
A plane brakes at (Ax) 10mi/h, after Vo of 160mi/h.

this gives acceleration, initial velocity, final velocity and time can be figured out.
 
  • #8
Do you mean changing km/h to m/s?
 
  • #9
Do you mean changing km/h to m/s?
 
  • #10
rculley1970 said:
Do you mean changing km/h to m/s?
No.
You got the right answer, but it is in [itex] m/s^2 [/itex], not on m/s
 
  • #11
I am taking a break for the night on it. Email if you can explain the hint a little bit more. Will be at work at it again in the morning. I know it isn't that hard of a problem but I am making it hard. I just need to figure out what I am doing wrong with the conversion for it. Thank you for your help.
 
  • #12
rculley1970 said:
I am taking a break for the night on it. Email if you can explain the hint a little bit more. Will be at work at it again in the morning. I know it isn't that hard of a problem but I am making it hard. I just need to figure out what I am doing wrong with the conversion for it. Thank you for your help.

? But you are done! You did find the acceleration!
 
  • #13
rculley1970 said:
I am taking a break for the night on it. Email if you can explain the hint a little bit more. Will be at work at it again in the morning. I know it isn't that hard of a problem but I am making it hard. I just need to figure out what I am doing wrong with the conversion for it. Thank you for your help.

you got ir right. You converted the distance and speed correctly. I was just pointing out that you gave your final answer with the wrong units. (but it's the correct numerical value)
 
  • #14
you mean -2.35 m/s^2. lol, i thought you meant I had the conversion wrong. Thank you for your help and I am going to redo the problem again in the morning just to verify. Sorry if I didn't get it soon enough but I guess I didn't carry the units across like I should have. I need to work on that. Once again, Thank you for all everyones help. By the way, this homework is already supposed to be submitted but I didn't get it done in time so I got a 0 on it. I am just trying to understand it because it may be on the exam coming up.
 
  • #15
rculley1970 said:
you mean -2.35 m/s^2. lol, i thought you meant I had the conversion wrong. Thank you for your help and I am going to redo the problem again in the morning just to verify. Sorry if I didn't get it soon enough but I guess I didn't carry the units across like I should have. I need to work on that. Once again, Thank you for all everyones help. By the way, this homework is already supposed to be submitted but I didn't get it done in time so I got a 0 on it. I am just trying to understand it because it may be on the exam coming up.

Sorry if I made you worry! :frown:

And you have the right attitude: it's very important to understand that very well in preparation for the tests.

good luck!
 

Related to Solve for Braking Acceleration for Emergency Stop

1. What is braking acceleration?

Braking acceleration is the rate at which an object slows down due to the application of brakes.

2. How is braking acceleration calculated?

Braking acceleration can be calculated by dividing the change in velocity by the time taken for the object to come to a complete stop. This can be represented by the equation a = (vf - vi) / t, where a is the braking acceleration, vf is the final velocity, vi is the initial velocity, and t is the time taken.

3. What factors affect the braking acceleration of a vehicle?

The braking acceleration of a vehicle is affected by factors such as the mass of the vehicle, the coefficient of friction between the tires and the road, and the efficiency of the brakes.

4. How does the braking acceleration for an emergency stop differ from a regular stop?

In an emergency stop, the braking acceleration will be higher as the driver will apply the brakes with more force in order to stop the vehicle as quickly as possible. This can also cause the vehicle to skid if not done carefully.

5. Can the braking acceleration be increased?

Yes, the braking acceleration can be increased by improving the efficiency of the brakes, maintaining proper tire pressure, and reducing the weight of the vehicle. However, it is important to note that exceeding the limits of braking acceleration can be dangerous and can cause the vehicle to lose control.

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