PhysicsLAB Resource Lesson
Average Velocity - A Calculus Approach

Motion of a particle in one dimension when given a position (displacement) function
 
Suppose a particle is moving back and forth along the x-axis with a position function (the coordinate giving the location of the particle on the x-axis) given by x(t) = t2 – 2t – 3
 
 
Assume that t is in seconds, t < 0 is not possible, and the coordinates on the x-axis are in meters. We already know the following facts:
 
  1. The initial position is x(0) = –3 meters which is 3 meters to the left of the origin
  2. The coordinate on the x-axis giving location when t = 2 is x(2) = 4 – 4 – 3 = –3 meters.
  3. The particle is at the origin when x(t) = 0 = (t – 3)(t + 1) or when t = 3 and t = –1
    (we exclude the case where t = –1 since we exclude t < 0).
  4. This position function x(t) is really a vector function called the "displacement" of the particle from the origin.
    x(0) = – 3 means the particle is a distance (magnitude of the vector x) of 3 to the left (direction) of the origin. 
 
Let's use these facts to answer the next two questions.
 
 What is the average velocity between times t = 2 and t = 5?

 What is the instantaneous velocity at t = 2?

 
Motion of a particle in one dimension when given a velocity or acceleration function
 
Suppose we know that the velocity of a particle moving back and forth on the x-axis is given by v(t) = 2t – 2 and the initial position x(0) = –3. Notice that this is the same particle as in Part I above.
 
 
Some basic facts:
 
  1. v(t) = 2t – 2 is the derivative of the position function x(t) = t2 – 2t – 3 .
  2. a(t) = 2 is the derivative of v(t) and is the uniform (or constant) acceleration of the particle.
  3. v(t) = 0 when t = 1 which invites us to investigate the sign of our velocity function
  4. v(t) < 0 when t < 1 and v(t) > 0 when t > 1.
 
To determine an object's position given its velocity function and an initial position we will use a basic fact from calculus:
 
Given the rate of change of a quantity and a time interval, the “net change” is the integral of the rate of change over the given time period.
 
 
Let's use these fact to look at three final questions.
 
 What is the position of the particle after the first 5 seconds of motion?

 What is the total distance traveled by the particle in the first 5 seconds?

 Is the particle speeding up or slowing down when t = 4?





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