Section 5: Motion

Elementary and Intermediate Algebra: Graphs & Models (3rd Edition)

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Section 5.5 Motion 505 Version: Fall 2007 5.5 Motion If a particle moves with uniform or constant acceleration, then it must behave according to certain standard laws of kinematics. In this section we will develop these laws of motion and apply them to a number of interesting applications. Uniform Speed If an object travels with uniform (constant) speed v , then the distance d traveled in time t is given by the formula d = vt, (1) or in words, “distance equals speed times time.” This concept is probably familiar to those of us who drive our cars on the highway. For example, if I drive my car at a constant speed of 50 miles per hour, in 3 hours I will travel 150 miles. That is, 150 mi = 50 mi h × 3 h Note that this computation has the form “distance equals speed times time.” It is important to note how the units balance on each side of this result. This is easily seen by canceling units much as you would cancel numbers with ordinary fractions. 150 mi = 50 mi h × 3 h In Figure 1 (a) we’ve plotted the speed v of the car versus time t . Because the speed is uniform (constant), the graph is a horizontal ray, starting at time t = 0 and moving to the right. In Figure 1 (b), we’ve shaded the area under the constant speed ray over the time interval [0 , 3] hours. Note that the area of the shaded rectangular region has height equal to 50 miles per hour ( 50 mi/h) and width equal to 3 hours ( 3 h), so the area of this rectangle is Area = height × width = 50 mi h × 3 h = 150 mi . Note the units on the answer. The area under the constant speed ray is 150 miles. That is, the area under the speed curve is the distance traveled! Our work has led us to the following result. Uniform Speed. Suppose that an object travels with uniform (constant) speed v . The distance traveled d is given by the formula d = vt , where t is the time of travel. The graph of speed v versus time t will be a horizontal ray, starting at time t = 0 and moving to the right. The area of the rectangular region under the graph of v over the time interval [0 , t ] gives the distance traveled during that time period.
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506 Chapter 5 Quadratic Functions Version: Fall 2007 t ( h ) v ( mi/h ) v =50 0 1 2 3 0 10 20 30 40 50 t ( h ) v ( mi/h ) v =50 0 1 2 3 0 10 20 30 40 50 50 mi/h 3 h Area =150 mi (a) Constant speed is represented by a horizontal ray. (b) The area under the speed curve is the distance traveled. Figure 1. Let’s look at another example. l⚏ Example 2. An object is traveling with uniform speed v . The graph of v versus t is shown in the graph that follows. t (s) v (ft/s) 0 10 20 0 100 v Figure 2. The graph of the speed of the object ver- sus time. What is the speed of the object at any time t ? How far will the object travel in 20 seconds? We read the speed from the graph. Note that the ray representing the speed is level (constant) at 100 feet per second ( 100 ft/s). Therefore, the speed at any time t is v = 100 . In function notation, we would write v ( t ) = 100 , being mindful that the units are feet per second (ft/s).
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Section 5.5 Motion 507 Version: Fall 2007 To find the distance traveled in 20 seconds, we have two choices: 1. If we use the formula d = vt , then d = vt d = 100 ft s × 20 s
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