# Homework Help: Ball rolling up an incline Final Exam Review

1. Dec 13, 2008

### charan1

1. The problem statement, all variables and given/known data
A ball rolls up a straight incline. The dots in the diagram represent the position of the ball every 4 seconds.
(The fourth dot is not intended to represent the turnaround point; the ball might still be on the way up at that
point.) The instantaneous velocities at the second and third dots are as given in the diagram.

a. Determine the velocities at the locations of the first and fourth dots and label them on the diagram. (2 points)

b. Determine how far the ball travels between the second and third dots. (3 points)

(IMAGE attached!)

2. Relevant equations

aavg= Delta V/ Delta T

3. The attempt at a solution
I am kind of stumped at this problem and not sure where to begin or how to get where I'm trying to go, all I could think to do was this...then I wasn't sure where to go from there.

(5m/s2 + 7m/s2) / 2 = Delta V/(8s-4s)

Delta V = 24m/s

#### Attached Files:

• ###### Sample Exam 1 Question 4 figure.jpg
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2. Dec 13, 2008

### charan1

woops! those are not accelerations they are velocities! sorry I will return when I attempt to figure it out again!

3. Dec 13, 2008

### charan1

Part C:
Determine the total time it takes the ball to roll up the incline, from the bottom to the turnaround point.

4. Dec 13, 2008

### charan1

Part A:
Equations:
Vf=Vi+a(delta T)
aavg=D=(delta V)/(delta T)

VDot 2=7m/s
VDot 3=5m/s

aavg=(5m/s-7m/s)/(8s-4s)=-.5m/s2

VDot 2=VDot 1+(-.5m/s/s)(4s-0s)
VDot 1=9m/s

VDot 4=VDot 3+(-.5m/s/s)(12s-8s)
VDot 4=3m/s

Part B:

Equation:
vf2=vi2+2a(delta x)

52=72+2(-.5m/s/s)(delta x)
(Delta x)=24m

Part C:

Equation:
vf=vi+a(delta T)

vf=0m/s
vi=9m/s
a=-.5m/s/s
(delta T)=?

0m/s=(9m/s)+(-.5m/s/s)(delta T)

(delta T)=18s