SOLUTION The equation \( d=17 t \) gives the distance a raccoon runs in feet as a function of time in seconds. The graph shows the distance a squirrel runs over time. Does the raccoon or the squirrel run faster? Show your work.

We’re given that the raccoon’s distance in feet is modeled by   d = 17t which tells us that the raccoon’s speed is 17 feet per second (since the coefficient of t is the speed). The graph (not shown here) plots how far the squirrel runs over time. To compare the speeds, we must determine the squirrel’s rate. This is done by finding the slope of the line in the squirrel’s distance vs. time graph. (Remember: slope = rise/run, which in a distance–time graph equals speed.) Suppose from the graph you can read two clear points on the squirrel’s line. For example, if the graph shows that at t = 0 seconds the squirrel is at 0 feet and at t = 6 seconds the squirrel is at 60 feet, then the slope (squirrel’s speed) would be calculated as   slope = (60 feet − 0 feet) / (6 seconds − 0 seconds) = 60/6 = 10 feet per second. Now compare the two speeds:  • Raccoon’s speed = 17 ft/sec  • Squirrel’s speed (from the graph) = 10 ft/sec Since 17 ft/sec is greater than 10 ft/sec, the raccoon runs faster than the squirrel. If your graph shows different points, use them to compute the slope of the line (the squirrel’s speed) in the same way. In every case, if the calculated slope is less than 17, then the raccoon is faster. Thus, based on the given equation and the information from the graph,  Answer: The raccoon runs faster.