How Far Will M1 Travel in 0.633 Seconds?

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The problem involves two masses, M1 and M2, with M1 being four times more massive than M2, and M1 is on a frictionless table while M2 hangs over a pulley. The calculated acceleration of M1 is 1.96 m/s², derived from the relationship a = (m2/(m1+m2))g. The velocity was determined to be 1.24 m/s using the formula a = v/t. However, the final distance calculation appears to be incorrect, as the relationship between velocity, time, and distance was misunderstood. The correct approach to find the distance traveled by M1 in 0.633 seconds needs clarification on the application of the formula for distance.
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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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