Antiderivatives - Examples, Part 4 - Position, Velocity, Acceleration Functions

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(Edited)

In this video I go further into the examples series on Antiderivatives and basically show how they can be used to determine the position function when given the velocity or the acceleration functions and also initial position and velocity values. I also discuss gravity and how near the surface of the Earth the acceleration due to gravity can be taken as a constant which is equal to 9.8 m/s2 or 32 ft/s2.

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Antiderivatives - Examples, Part 4 - Position, Velocity, Acceleration Functions

AntiDerivatives Examples Part 4.jpeg

Example 1

A particle moves in a straight line and has acceleration given by a(t) = 6t + 4. Its initial velocity is v(0) = -6 cm/s and its initial displacement is s(0) = 9 cm. find its position function s(t).

Solution:

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Example 2

A ball is thrown upward with a speed of 48 ft/s from the edge of a cliff 432 ft above the ground. Find its height above the ground t seconds later. When does it reach its maximum height? When does it hit the ground?

  • An object near the surface of the earth is subject to a gravitational force that produces downward acceleration denoted by g.
  • For motion close to the ground we may assume that g is constant, its value being 9.8 m/s2 (or 32 ft/s2)

Solution:

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