2.1.2 The Uniform-Motion Model

Chapter 1 introduced a model as a simplified picture of reality, but one that still captures the essence of what we want to study. When it comes to motion, few real objects move with a precisely constant velocity. Even so, there are many cases in which it is quite reasonable to model their motion as being uniform. That is, uniform motion is a very good approximation of their actual, but more complex, motion. The uniform-motion model is a coherent set of representations—words, pictures, graphs, and equations—that allows us to explain an object’s motion and to predict where the object will be at a future instant of time.

MODEL 2.1 Uniform motion

For motion with constant velocity.

  • Model the object as a particle moving in a straight line at constant speed:

    The figure illustrates a motion diagram and its velocity-versus-time and position-versus-time graphs to explain the uniform motion model.
  • Mathematically:

    • vs=Δs/Δt

    • sf=si+vs Δt

  • Limitations: Model fails if the particle has a significant change of speed or direction.

Workbook Exercise 4

EXAMPLE 2.2 Lunch in Cleveland?

Bob leaves home in Chicago at 9:00 a.m. and drives east at 60 mph. Susan, 400 miles to the east in Pittsburgh, leaves at the same time and travels west at 40 mph. Where will they meet for lunch?

MODEL Here is a problem where, for the first time, we can really put all four aspects of our problem-solving strategy into play. To begin, we’ll model Bob’s and Susan’s cars as being in uniform motion. Their real motion is certainly more complex, but over a long drive it’s reasonable to approximate their motion as constant speed along a straight line.

VISUALIZE FIGURE 2.5 shows the pictorial representation. The equal spacings of the dots in the motion diagram indicate that the motion is uniform. In evaluating the given information, we recognize that the starting time of 9:00 a.m. is not relevant to the problem. Consequently, the initial time is chosen as simply t0=0 h. Bob and Susan are traveling in opposite directions, hence one of the velocities must be a negative number. We have chosen a coordinate system in which Bob starts at the origin and moves to the right (east) while Susan is moving to the left (west). Thus Susan has the negative velocity. Notice how we’ve assigned position, velocity, and time symbols to each point in the motion. Pay special attention to how subscripts are used to distinguish different points in the problem and to distinguish Bob’s symbols from Susan’s.

FIGURE 2.5 Pictorial representation for Example 2.2.

A pictorial representation of the motion of two vehicles traveling toward each other from opposite directions.

One purpose of the pictorial representation is to establish what we need to find. Bob and Susan meet when they have the same position at the same time t1. Thus we want to find (x1)B at the time when (x1)B=(x1)S. Notice that (x1)B and (x1)S are Bob’s and Susan’s positions, which are equal when they meet, not the distances they have traveled.

SOLVE The goal of the mathematical representation is to proceed from the pictorial representation to a mathematical solution of the problem. We can begin by using Equation 2.3 to find Bob’s and Susan’s positions at time t1 when they meet:

(x1)B=(x0)B+(vx)B (t1−t0)=(vx)B t1(x1)S=(x0)S+(vx)S (t1−t0)=(x0)S + (vx)S t1

Notice two things. First, we started by writing the full statement of Equation 2.3. Only then did we simplify by dropping those terms known to be zero. You’re less likely to make accidental errors if you follow this procedure. Second, we replaced the generic symbol s with the specific horizontal-position symbol x, and we replaced the generic subscripts i and f with the specific symbols 0 and 1 that we defined in the pictorial representation. This is also good problem-solving technique.

The condition that Bob and Susan meet is

(x1)B=(x1)S

By equating the right-hand sides of the above equations, we get

(vx)B t1=(x0)S+(vx)S t1

Solving for t1 we find that they meet at time

t1=(x0)S(vx)B−(vx)S=400 miles60 mph−(−40) mph=4.0 hours

Finally, inserting this time back into the equation for (x1)B gives

(x1)B=(60mileshour)×(4.0 hours)=240 miles

As noted in Chapter 1, this textbook will assume that all data are good to at least two significant figures, even when one of those is a trailing zero. So 400 miles, 60 mph, and 40 mph each have two significant figures, and consequently we’ve calculated results to two significant figures.

While 240 miles is a number, it is not yet the answer to the question. The phrase “240 miles” by itself does not say anything meaningful. Because this is the value of Bob’s position, and Bob was driving east, the answer to the question is, “They meet 240 miles east of Chicago.”

REVIEW Before stopping, we should check whether or not this answer seems reasonable. We certainly expected an answer between 0 miles and 400 miles. We also know that Bob is driving faster than Susan, so we expect that their meeting point will be more than halfway from Chicago to Pittsburgh. Our review tells us that 240 miles is a reasonable answer.

It is instructive to look at this example from a graphical perspective. FIGURE 2.6 shows position-versus-time graphs for Bob and Susan. Notice the negative slope for Susan’s graph, indicating her negative velocity. The point of interest is the intersection of the two lines; this is where Bob and Susan have the same position at the same time. Our method of solution, in which we equated (x1)B and (x1)S, is really just solving the mathematical problem of finding the intersection of two lines. This procedure is useful for many problems in which there are two moving objects.

FIGURE 2.6 Position-versus-time graphs for Bob and Susan.

The figure illustrates a position-versus-time graph drawn on a coordinate plane.