Laws of motion magnitude of average acceleration

[maxiJ]

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.

 

[Hootenanny]

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?

 

[maxiJ]

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.

 

[Hootenanny]

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?

 

[maxiJ]

ah, yes.. i overlooked that.

well for average velocity its change in displacement over change in time, i think.

 

[Hootenanny]

well for average velocity its change in displacement over change in time, i think.

Whilst that is correct, I think that you may find some more useful formulae on https://www.physicsforums.com/showpost.php?p=905663&postcount=2".

 

[maxiJ]

wow that was a lot easier than i made it out to be. thanks for the help.

 

[Hootenanny]

wow that was a lot easier than i made it out to be. thanks for the help.

My pleasure.