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A point mass moving on a vertical line has an acceleration given by the equation a = 10t – 3. At t = 0, the displacement is found to bes = - 5m. And after at t = 7s, the displacement is s = 12m. - %3D %3D Derive the equation of its motion. Use the boundary conditions to solve for the constants of integration.

A point mass moving on a vertical line has an acceleration given by the equation a = 10t – 3. At t = 0, the displacement is found to bes = - 5m. And after at t = 7s, the displacement is s = 12m. - %3D %3D Derive the equation of its motion. Use the boundary conditions to solve for the constants of integration.

College Physics
College Physics
1st Edition
ISBN:9781938168000
Author:Paul Peter Urone, Roger Hinrichs
Publisher:Paul Peter Urone, Roger Hinrichs
Chapter1: Introduction: The Nature Of Science And Physics
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A point mass moving on a vertical line has an acceleration given by the equation a = 10t – 3. At t = 0, the displacement is found to be s = - 5m. And after at t = 7s, the displacement is s = 12m. Derive the equation of its motion. Use the boundary conditions to solve for the constants of integration.
Transcribed Image Text:A point mass moving on a vertical line has an acceleration given by the equation a = 10t – 3. At t = 0, the displacement is found to be s = - 5m. And after at t = 7s, the displacement is s = 12m. Derive the equation of its motion. Use the boundary conditions to solve for the constants of integration.
C1 C2 eqn.
Transcribed Image Text:C1 C2 eqn.
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