Does a Ball Move Up or Down an Inclined Plane?

[alwysnforevr002]

Is the motion of a ball on an inclined plane mean the ball is going up the inclined plane or down the inclined plane?

The question goes as follows

The motion of a ball on an inclined plane is described by the equation . Which of the following quantities must have a value of zero?
xi
xf
vi
tf

 

[alwysnforevr002]

Okay let me start over. This question is confuzing me.
If i can figure out what it means than i already know the answer. I am simply looking for a further explanation. Basically the question is

The motion of a ball on an inclined plane is described by the equation DeltaX = 1/2(acceleration(delta t)(squared))
Which of the following quantities must have a value of zero?
xi
xf
vi
tf


NOW, my problem is that i don't know if the "inclined plane" is a plane starting from the top and then going down or starting from the bottom and going up.

I hope that is bit more clear. i apologize for the mess

 

[berkeman]

The question is poorly presented to you, if that is the exact wording. Start the ball at the top of the inclined plane with some initial velocity, then nothing is zero. Ask you professor for a better problem statement. But be polite when you ask, unless you are already an A student aceing the class. Then poke some fun at him/her.

 

[berkeman]

Which of the following quantities must have a value of zero?
xi
xf
vi
tf

And even if the ball is started up the inclined plane with some initial velocity, none of the quantities above would be zero. The only zero would come from when the ball's velocity slowed to zero and reversed itself due to gravity (easiest calc is PE(zero velocity) = KE(initial)).

 

[alwysnforevr002]

Thank you. I will make sure to look that up. For now I have to pick the best answer and i think that would be initial velocity.

 

[berkeman]

Thank you. I will make sure to look that up. For now I have to pick the best answer and i think that would be initial velocity.

Like if it starts at rest on an inclined plane? Hmmm, that would seem to work. I didn't think of that.