Solve Mechanics Problem: ax=0.5m/s^2, 9 & 15m Inclines, v1=0m/s

[G. Coder]

Homework Statement

ax=0.500 m/s^2
Incline plane=9 m
second incline plane=15m
v1=0m/s(object starts at rest
t(time)=?
Problem also is uploaded

Homework Equations

vf=vi+axt

The Attempt at a Solution

For part A.[/B]
1. Since ax=0.5m/s^2,
2. .5(m/s^2)/9m=.055555551/s^2
3. Invert: (.055555551/s^2)^-1=18s^2
4. sqrt: 18s^2=4.24 s
5. Plug in: Get 2.1 m/sThe answer is supposed to be 3.00 m/s. Is there suppose to be a different way of setting it up.

 

[Nathanael]

For part A.
1. Since ax=0.5m/s^2,
2. .5(m/s^2)/9m=.055555551/s^2
3. Invert: (.055555551/s^2)^-1=18s^2
4. sqrt: 18s^2=4.24 s
5. Plug in: Get 2.1 m/s

Although you made the dimensions correct, this is not a valid way of finding the final speed.
What is your interpretation of the acceleration divided by the distance? (It seems rather meaningless to me)

This is how I would do it:
Call the unknown final speed "V"
In terms of V, (and the given acceleration) how long will it take to reach the bottom?
In terms of V, what is the average velocity of the ball?
If you multiply the average velocity and the time it took, what should the answer come out to be?

 

[haruspex]

What is your interpretation of the acceleration divided by the distance? (It seems rather meaningless to me)

Actually it is valid, and arises from a standard SUVAT equation. However, there's a constant factor missing. G. Coder, what equation did you base that on? It's not the one you quoted.