Calculating Acceleration and Distance in Kinematic Equations

In summary, the conversation is about a physics question on finding the acceleration and distance a train travels in a given time. The student is having trouble solving for the distance using the equation x= 1/2(final vel + initial vel)t and x= final vel(time) + 1/2at squared. With the help of others, the student realizes they should be using the equation x= initial vel*time + 1/2at squared and the initial velocity should be converted to meters per second. After making these corrections, the student is still having trouble getting the correct answer and seeks clarification.
  • #1
toddler
17
0
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
 
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  • #2
X=Final Vel(time) + 1/2at squared
you didn't copy this equation correctly
 
  • #3
sorry, I meant

x= initial vel*time +1/2at squared
 
  • #4
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?
 
  • #5
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
 
  • #6
You should be using 25000m/3600s for your initial velocity, not 25. That'll get you 106. :)
 
  • #7
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:
 
  • #8
ahhh thank you everyone, really appreciate it
 
  • #9
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
 

FAQ: Calculating Acceleration and Distance in Kinematic Equations

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