Question is about the acceleration

[gijungkim]

Homework Statement

The system shown is released from rest and moves 50 cm in 1.0 s. What is the value of M? All surfaces are frictionless.

X = 0.5 m
t = 1sec
m1 = 3kg
m2 = M

However, my question is about the acceleration. I'm just curious why I cannot use vf = v0 + at equation instead of x = x0 + v0t + 0.5at^2

Homework Equations

x = x0 + v0t + 0.5at^2
vf = v0 + at

The Attempt at a Solution

X = X0 + V0t + 0.5at^2
0.5m = 0 + 0 + 0.5a(1)^2
a=1m/s^2

M*g = (3kg + M)a
M(g-a) = 3a
M = 3a/(g-a) = 0.34kg

So I got M = 0.34kg. I know this is the answer but now my question is the acceleration.
I know I got 1m/s^2 from X = X0 + V0t + 0.5at^2 equation, but why can I not use Vf = V0 + at equation?
If I use Vf = V0 + at then I get a=0.5m/s^2 since vf = 0.5m/s and v0 = 0m/s and t=1.
Why do they have different acceleration?

 

[Chestermiller]

Where did you get Vf = 0.5 m/s from?

Chet

 

[gijungkim]

Where did you get Vf = 0.5 m/s from?

Chet

Since it moved 0.5m in 1 sec, isn't it 0.5m/s??

 

[Chestermiller]

Since it moved 0.5m in 1 sec, isn't it 0.5m/s??

That's the average velocity, not the final velocity. The velocity at time zero is 0 m/s and the final velocity at the end of 1 sec is 1m/s. So, what's the average of these two?

Chet

 

[gijungkim]

That's the average velocity, not the final velocity. The velocity at time zero is 0 m/s and the final velocity at the end of 1 sec is 1m/s. So, what's the average of these two?

Chet

0.5! Thank you so much :)