Motion at Constant Acceleration Question

AI Thread Summary
A car accelerates from 13 m/s to 25 m/s over 6.0 seconds, resulting in an acceleration of 2.0 m/s². To find the distance traveled during this time, the kinematic equation x = x0 + v0t + 1/2 at² can be applied. The user confirms they have solved for distance using this equation. They inquire whether to continue seeking help in the same thread or to create a new one for further questions. The discussion emphasizes the importance of kinematic equations in solving motion problems under constant acceleration.
Adrianna
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Homework Statement


A car accelerates from 13 m/s to 25 m/s in 6.0s. What was its acceleration? How far did it travel in this time? Assume constant acceleration.


Homework Equations


I used the equation a=v/t


The Attempt at a Solution


I got that the acceleration is 2.0 m/s^2. Now my problem in that I am not sure how to go about finding the distance it went in the time.
 
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Does your book give the kinematic equations?
 
well it gives me
v=v0 +at
x=x0 + v0t+ 1/2 at^2
v^2 = v^20 + 2a(x -x0)
v= (v+v0)/2
 
v0 is initial velocity, x0 is initial position, and x is final position. How far something travels is the difference between its final and initial positions. Using the second equation should help.
 
Okay I got that one thanks so should I post another thread to get help on the next one or should I just ask it here?
 
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