Constant slope acceleration problem

[shuvit760]

I got this problem:

A ball is rolled up a constant slope with an initial velocity of 11.0 m/s and after 9.3 s the ball is rolling down the slope with a velocity of 7.3 m/s. What is the acceleration of the ball on the slope.

Here's what I tried: I divided the problem in 2 for the ball going up, and the ball going down.

1st Givens: Vi: 11 m/s Vf: 0 t:9.3
2nd Givens: Vi: 7.3 m/s Vf :0

I then got the displacement of half of the slope from the first set of givens, which gave me 51.15. That's where I got stuck. I tried getting a acceleration for both sides of the slope and then averaging, but that wasn't right.

Please help

 

[kreil]

$$a=\frac{{\Delta}v}{t} \implies a=\frac{v_f-v_i}{t}$$

**hint** velocity has direction, so when the ball is rolling down the slope, it has a negative velocity.

.

 

[shuvit760]

Ah yes, but there is still a problem. This is what changes then:
2nd Givens: Vi: -7.3 m/s D:-51.15 Vf: 0(I'm not sure if this is right)
So I can get a time for when it's going down of 14 sec.
So I added the inicial velocities of the ball going up and going down and then divided it by 23.3(9.3 sec + 14 sec). But it gave a large decimal.

The teacher gave us the answer of -2.0 m/s^2.

Damn, I'm still stuck. Please advise.

 

[kreil]

The problem tells you that it starts at 11m/s, and 9.3s later it is going down at -7.3 m/s. just plug this information into the equation:

$$a=\frac{-7.3m/s-11m/s}{9.3s}=\frac{-18.3}{9.3}=-1.96m/{s^2}$$

 

[shuvit760]

Wow, I just complicating myself with all my equations. Thank you so much for your help