Calculating Acceleration and Distance in Kinematic Equations

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Homework Help Overview

The discussion revolves around kinematic equations in the context of a train accelerating from an initial velocity to a final velocity over a specified time period. The original poster seeks to calculate both the acceleration and the distance traveled by the train during this time.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate acceleration and distance using various kinematic equations but encounters discrepancies with the expected results. Some participants question the application of the equations and suggest corrections to the initial conditions used in the calculations.

Discussion Status

Participants are actively engaging with the original poster's calculations, offering guidance on the correct use of units and equations. There is a focus on clarifying the setup of the problem and ensuring the correct values are applied in the equations. Multiple interpretations of the equations are being explored, but no consensus has been reached on the final calculations.

Contextual Notes

There is a noted confusion regarding the conversion of units from km/hr to m/s, which affects the calculations. The original poster expresses frustration over the discrepancies between their results and those found in the textbook.

toddler
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1Dim. Physics question

I'm in my 3rd weeks of physics, so I'm dealing with Kinematic Equations.

My question is: A train accelerates from an initial velocity of 25.0 km/hr to a final vel. of 65.0km/h in 8.5 seconds..Find its acceleration and the distance the train travels during this time.

for acceleration, I did A = Final Vel - Initial Vel / time and when put in m/s , it comes out to 1.31 meters per second squared...i checked the answered in the back of the book..it checks out..ok, i got that part...

now... for distance, I used X = 1/2(final vel + initial Vel)t no matter how many times I did it, my answer doesn't match up with the books correct answer of 106m

i also thought i might have used the wrong equation so i tried using:

X=Final Vel(time) + 1/2at squared

still doesn't come out right


what am i doing wrong? it's really frustrating, ...I need to get this and do real well because its part of my pre-med requirement, so any help would be appreciated..thanks
 
Last edited:
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X=Final Vel(time) + 1/2at squared
you didn't copy this equation correctly
 
sorry, I meant

x= initial vel*time +1/2at squared
 
ok, and you used this equation and it did not work?
can you write out the exact equation (with the numbers you used) to solve for the second part?
 
x= initial velocity(time) + 1/2at squared

x = 25(8.5) + 1/2(1.31)(8.5 squared)
x= 212.5 + .655(8.5squared)
x= 212.5 + 72.91
x= 285.41

yet, x is supposed to be 106 in the book
 
You should be using 25000m/3600s for your initial velocity, not 25. That'll get you 106. :)
 
toddler said:
sorry, I meant

x= initial vel*time +1/2at squared

x = initial position + initial vel*time + 1/2at squared. Irrelevant in this problem, but somewhere else it just may not be. :smile:
 
ahhh thank you everyone, really appreciate it
 
damn...i used Oksanav's advice and here's where I am stuck


x= 1/2 (25,000m/3,600s + 65,000m/3,600s)(8.5s)

x= 1/2 (90,000m/7,200s) (8.5s)

now here's where I am gettin stuck...im multiplying (90,000m/7,200s) by (8.5s) and getting 106.25 and then taking half of that because of the '1/2' at the beginning of the equation...the answer in the book is 106 though..so what did i do wrong

x= 1/2 (106.25) = 53.125??
 
  • #10
Hello toddler,

you are not applying the correct equation in your last post.

You've already found out in post #5 that:

[tex]x=v_{initial}t+\frac{1}{2}at^2[/tex]

with a being

[tex]a=\frac{(v_{final}-v_{initial})}{t}[/tex]

Regards,

nazzard
 

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