Calculating Acceleration of a Braking Train

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Homework Help Overview

The discussion revolves around calculating the average acceleration of a train that decelerates from a speed of 140 km/h to a stop over a distance of 700 meters. The subject area includes kinematics and the application of equations of motion.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore different kinematic equations to determine the average acceleration, questioning the appropriateness of the formulas used. There is discussion about the conditions under which certain equations apply, particularly regarding constant velocity versus changing velocity.

Discussion Status

Some participants have identified potential equations to use and have calculated values for acceleration. There is an ongoing exploration of how to confirm the accuracy of these calculations, indicating a productive direction in the discussion.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the information they can reference or the methods they can employ. There is also a focus on ensuring the correct application of kinematic equations in the context of acceleration and deceleration.

Nanu Nana
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Homework Statement


A train that runs 140 km / h requires about 700m to stop. Calculate the average acceleration of the train while braking.

Homework Equations


v: x/t[/B]

The Attempt at a Solution


v= x/t
so t= x/v= 700m/ (38.89m/s)=18 s
Now to find acceleration a= v/t
a= 38,89m/s / 18s =2,160 m/s^2
They're asking the average so 2,160 m/s2 is our initial acceleration
therefore
0-2,160 m/s^2 / (2) = -1,08024
Is it the correct formula to use ? or is it wrong ?
 
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The equation you used for the velocity is correct only if the acceleration is zero (i.e., only if the velocity is constant).
 
Oh I see. I think I've found out which kinematic equation to use .

I tried this one
initial velocity = 140 km/h (38,89 m/s)
Final velocity = 0 m/s because its going to stop
(vx)^2=(u)^2 + 2x a x700 m
(38,89)^2=1400 x a
a= -1512,34/ 1400
a= -1,0802
 
Nanu Nana said:
Oh I see. I think I've found out which kinematic equation to use .

I tried this one
initial velocity = 140 km/h (38,89 m/s)
Final velocity = 0 m/s because its going to stop
(vx)^2=(u)^2 + 2x a x700 m
(38,89)^2=1400 x a
a= -1512,34/ 1400
a= -1,0802

That's correct. Can you think of a way to confirm that answer is right?
 
My workbook says so
 
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