Ball rolling up an incline Final Exam Review

AI Thread Summary
The discussion revolves around a physics problem involving a ball rolling up an incline, focusing on calculating velocities and distances at specific time intervals. The user initially struggles with identifying the correct approach but later provides calculations for velocities at the first and fourth dots, determining them to be 9 m/s and 3 m/s, respectively. They also calculate the distance traveled between the second and third dots as 24 m. Additionally, the total time taken for the ball to reach the turnaround point is found to be 18 seconds. The user seeks verification of their calculations and understanding of the concepts involved.
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Homework Statement


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!)


Homework Equations



aavg= Delta V/ Delta T

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

Please help! thank you
 

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

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

Please check thank you!
 
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