What are the calculations for a ball rolling up an incline plane?

AI Thread Summary
The discussion focuses on calculating the motion of a ball rolling up an incline. For part A, the velocities at the first and fourth dots were determined to be 9 m/s and 3 m/s, respectively, using the average acceleration of -0.5 m/s². In part B, the distance traveled between the second and third dots was calculated to be 24 meters. For part C, the total time for the ball to reach the turnaround point was found to be 18 seconds. The calculations appear to be correct based on the provided data.
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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)

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


The Attempt at a Solution



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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