How Far Will M1 Travel in 0.633 Seconds?

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SUMMARY

The problem involves two blocks, M1 and M2, where M1 is four times more massive than M2, and M1 is on a frictionless table while M2 hangs over a pulley. The acceleration of the system is calculated as a = (1/5)g, resulting in a value of 1.96 m/s². The velocity is derived using the formula a = v/t, yielding a velocity of 1.24 m/s. The distance traveled by M1 in 0.633 seconds is calculated as 0.785 meters, but the user receives an incorrect answer from their homework system, indicating a misunderstanding in the application of the distance formula.

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The problem is this: There are two blocks of unknown masses M1 and M2. M1 is sitting on a frictionless table and M2 is attached to m1 by a massless string and is hanging over the edge over a pulley. If m1 is four times more massive than m2, how far will m1 travel in 0.633 seconds?
So far I've gotten
a= (m2/m1+m2)g .
I replaced m1 with 4*m2, and then simplified the equation to give me
a=(1/5)g
a=(1/5) 9.8
a= 1.96 m/s^2.

Once I had the acceleration, I then used the fact that a=v/t to give me the velocity, and then I multiplied the velocity by the time to give me distance.
a=v/t
1.96=v/.633 sec
v=1.24m/s
I then multiplied that by the time in order to give me the distance.
v*t= 0.785 m

I entered this into the homework on the internet and it says I'm wrong, I don't know what else to do? Can anyone help me on where I messed up?
thanks
 
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it's not a=v/t, but a=dv/dt.

you did solve for v correctly and a correctly though and you almost have the right answer. the error is in the last part of your work.

what does (velocity)(time)=distance mean?
 

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