Acceleration Problem: Solving for Constant Acceleration with Kinematic Equations

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The discussion revolves around understanding displacement and acceleration in kinematic equations, particularly in the context of a car slowing down from 23 m/s to rest over 85 m. The original poster expresses confusion about determining the direction of displacement and acceleration when not specified in the problem. They calculated the acceleration as -3.1 m/s², which is correct, as it indicates deceleration opposite to the initial velocity direction. Clarification is provided that the acceleration must be negative due to the car's slowing down. The conversation highlights the importance of understanding vector directions in kinematics.
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Homework Statement



I'd just like someone to see if I answered correctly. I'm having a bit of difficulty grasping the vagueness of the displacement in these questions I've been getting in my textbook. From what I gather the displacement is relative to the vector points, (the two x points in kinematic equations). However, how am I to know what direction the car is traveling relative to the x plane if it doesn't specify the point in a question? I don't know if you guys get me.. The acceleration I found in the question below, could be positive or negative, it doesn't specify enough. Or am I wrong?

A car slows down from 23 m/s to rest in a distance of 85 m. What was its acceleration, assumed constant?

Homework Equations



The kinematic equations for constant acceleration.

I attempted to solve it with the following:

finalV^2=initialV^2+2a(finalX-initialX)

The Attempt at a Solution



Here is an image of my attempted work. I got -3.1 m/s^2 as my answer.
Physics_problem_solution.jpg
 
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Good work by you! :smile:
 
Yes! Thanks
 
The acceleration I found in the question below, could be positive or negative, it doesn't specify enough. Or am I wrong?

A car slows down from 23 m/s to rest in a distance of 85 m. What was its acceleration, assumed constant?

Sorry, I did not read your question carefully enough.

V0x = +23m/s (assumedly in the +x direction)
x0 = 0
x = 85

Since the car is decelerating, the acceleration vector will be in the opposite direction of the velocity vector, hence the appropriateness of the "-" in front of 3.1m/s2
 
Very helpful, thank you.
 
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