Laws of motion magnitude of average acceleration

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Homework Help Overview

The problem involves calculating the magnitude of average acceleration for a rock that starts at a speed of 5.0 m/s, travels a distance of 3.0 m, and then comes to a stop. The context is within the subject area of kinematics.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the method of calculating time based on initial speed and distance, with one participant questioning the validity of their approach. There is an exploration of kinematic equations and clarification of terms like average velocity versus initial velocity.

Discussion Status

The discussion is ongoing, with participants providing guidance and questioning assumptions. There is recognition of a misunderstanding regarding the relationship between initial velocity and average velocity, and some participants suggest looking into additional kinematic equations.

Contextual Notes

Participants are navigating through the definitions of average velocity and acceleration, with some confusion regarding the correct application of formulas. There is an acknowledgment that the initial velocity is not the average velocity, which is a key point of discussion.

maxiJ
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Homework Statement


A rock is rolled in the sand. It starts at 5.0 m/s, moves in a straight line for a distance of 3.0m and then stops. What is the magnitude of the average acceleration?

Homework Equations


aavg = change in V/ change in time
magnitude of acceleration = change of speed/time interval.

The Attempt at a Solution


i think i am thinking too hard about this problem. seems simple but cannot get the right answer
found time = 1.67 s and tried to plug in for average acceleration. 5.0/1.67 = 2.99 m/s^2
the correct answer is 4.2 m/s^2.
?? can't figure out what i am doing wrong.
 
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maxiJ said:

Homework Statement


A rock is rolled in the sand. It starts at 5.0 m/s, moves in a straight line for a distance of 3.0m and then stops. What is the magnitude of the average acceleration?

Homework Equations


aavg = change in V/ change in time
magnitude of acceleration = change of speed/time interval.

The Attempt at a Solution


i think i am thinking too hard about this problem. seems simple but cannot get the right answer
found time = 1.67 s and tried to plug in for average acceleration. 5.0/1.67 = 2.99 m/s^2
the correct answer is 4.2 m/s^2.
?? can't figure out what i am doing wrong.
Welcome to Physics Forums,

Tell me, how did you find the time?
 
thanks =)

i took 5.0 m/s and divided by the distance of 3.0 m to get 1.67 seconds.
so the ball traveled 3.0 m in 1.67 s at 5.0 m/s.
 
maxiJ said:
thanks =)

i took 5.0 m/s and divided by the distance of 3.0 m to get 1.67 seconds.
so the ball traveled 3.0 m in 1.67 s at 5.0 m/s.
I'm afraid that's incorrect, it would be correct if 5.0m/s was the average velocity, but it isn't it's the initial velocity.

Do you know any kinematics equations?
 
ah, yes.. i overlooked that.

well for average velocity its change in displacement over change in time, i think.
 
Last edited by a moderator:
wow that was a lot easier than i made it out to be. thanks for the help.
 
maxiJ said:
wow that was a lot easier than i made it out to be. thanks for the help.
My pleasure.
 

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