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Finding the Distances Traveled by Moving Particles on Lines

Instructor: Matthew Bergstresser

Matthew has a Master of Arts degree in Physics Education. He has taught high school chemistry and physics for 14 years.

How much distance a moving object covers depends on how fast it is going and/or accelerating and for how long. In this lesson, we will use integral Calculus to determine the distance moving objects travel.

Motion Along Straight Lines

Think of all of the objects that move in straight lines. There are a lot of them, aren't there? Bugs can move in straight lines, humans run in straight lines away from bugs, and cars move in straight lines among many other things. Let's focus on the motion of a few objects moving in a straight line and see how we can use integral calculus to determine the distance it moves.

Distance versus Displacement

Imagine a person walking 5 meters to the right, and then moving 5 meters to the left as depicted in Diagram 1.


Diagram 1. A person moves 5 m to the left and then 5 m to the right.
d2


Since the person started and stopped at the same location their displacement is 0 meters. Mathematically written this is +5 m + (-5 m) = 0 m. Displacement is the length of a straight line drawn from where the object starts moving to where it stops moving. The person certainly moved, though and we represent the distance they moved as the absolute value of these displacements. Mathematically written this is |5 m| + |-5 m| = 10 m. Now let's see how these concepts translate into graphs of velocity versus time.

Velocity versus Time

Linear Graph Segments

To know how far an object has moved we need to know its velocity and how long it has been moving. Diagram 2 shows the velocity of an object with respect to time and we will determine the distance it moves from 0 to 12 seconds.


Diagram 2. Velocity versus time.
g1


We can see from this graph it is composed of three linear segments. The area between the function and the time-axis is displacement, which is known as the integral of the function. Seeing that the graph dips below the time-axis the displacement will include negative values. This means the displacement and distance are not going to be equal. To get the distance the object travels we need to determine the area between the function and the time axis and we need to take the absolute value of the areas. Let's calculate it now and we'll start by defining the areas we need to calculate shown in Diagram 3.


Diagram 3. We have to calculate 4 areas and sum their absolute values.
g1areas


We have right triangles and rectangles as the areas so we will use their area equations. We'll set up a chart to keep the areas organized. Recall that we are taking the absolute values of each area.

Distance Traveled
Areas: A1 A2 A3 A4
Time segments: 0 - 5 s 5 - 8 s 8 - 10 s 10 - 12 s
Equations: (1/2)bh bh (1/2)bh (1/2)bh
Plugging in values: (1/2)(5 s)(3 m/s) (3 s)(3 m/s) (1/2)(2 s)(3 m/s) (1/2)(2 s)(-5 m/s)
Distances: 7.5 m 9 m 3 m 5 m

Total distance 24.5 m

Let's now deal with a scenario when we can't use geometric formulas to determine the area between the function and the time-axis.

Non-linear Graphs

Let's say we have an object that is initially moving slowly and gradually accelerates and reaches a maximum velocity before slowing down fairly quickly. The object then moves in the other direction accelerating at a fairly constant rate eventually slowing down to rest. Maybe this depicts the motion of a person being chased by a bee! We can model the person's velocity as v = t5 -3t4 +2t2. The graph of this function is shown in Diagram 4.


Diagram 4. Running away from a bee.
graph2a


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