Laws of motion magnitude of average acceleration

AI Thread Summary
The discussion revolves around calculating the magnitude of average acceleration for a rock that starts at 5.0 m/s, travels 3.0 m, and stops. The initial attempt to find the time was incorrect, as the user divided the initial speed by distance instead of using kinematic equations. Correctly applying the kinematic equations reveals that the average acceleration is 4.2 m/s², not the initially calculated 2.99 m/s². The conversation highlights the importance of distinguishing between initial velocity and average velocity in motion problems. Ultimately, the user realizes the solution is simpler than initially thought with the right approach.
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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