Solve for Braking Acceleration for Emergency Stop

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SUMMARY

The discussion centers on calculating the braking acceleration of a train traveling at 297 km/h that requires 1.45 km to come to a complete stop. The relevant formula used is \( v_f^2 = v_i^2 + 2a \Delta x \), where \( v_f \) is the final velocity (0 m/s), \( v_i \) is the initial velocity (82.5 m/s after conversion), and \( \Delta x \) is the distance (1450 m). The calculated acceleration is -2.35 m/s², indicating deceleration. The importance of unit conversion and understanding the context of acceleration in horizontal motion is emphasized.

PREREQUISITES
  • Understanding of kinematic equations, specifically \( v_f^2 = v_i^2 + 2a \Delta x \)
  • Ability to convert units from km/h to m/s
  • Basic knowledge of acceleration and deceleration concepts
  • Familiarity with the implications of negative acceleration in motion
NEXT STEPS
  • Practice solving kinematic equations with different initial and final velocities
  • Learn more about unit conversions between different measurement systems
  • Explore real-world applications of acceleration in transportation systems
  • Study the effects of friction and other forces on braking distances
USEFUL FOR

Students studying physics, engineers involved in transportation design, and anyone interested in understanding motion dynamics and braking systems.

rculley1970
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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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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!
v_f^2 = v_i^2 + 2 a_x \Delta x [/tex]! That's all you need!
 
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?
 
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.
 
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.
 
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...
 
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.
 
Do you mean changing km/h to m/s?
 
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 m/s^2, 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!
 

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