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

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SUMMARY

The discussion focuses on the calculations involved in analyzing a ball rolling up an incline plane, specifically addressing the determination of velocities at various points and the distance traveled. Using the equations of motion, the initial velocity at the first dot is calculated as 9 m/s, and the velocity at the fourth dot is found to be 3 m/s. The distance traveled between the second and third dots is determined to be 24 meters, and the total time taken to reach the turnaround point is calculated to be 18 seconds.

PREREQUISITES
  • Understanding of kinematic equations, specifically Vf = Vi + a(delta T)
  • Knowledge of average acceleration calculations, aavg = (delta V)/(delta T)
  • Familiarity with the concept of instantaneous velocity
  • Ability to interpret and analyze motion diagrams
NEXT STEPS
  • Study the application of kinematic equations in different motion scenarios
  • Learn about the effects of friction on rolling motion
  • Explore advanced topics in dynamics, such as energy conservation in rolling objects
  • Investigate the impact of incline angles on rolling motion calculations
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and motion analysis, as well as educators looking for examples of kinematic applications in real-world scenarios.

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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

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