the distance, in feet, two boys travel per second on a treadmill is shown to the left. which comparison is…

the distance, in feet, two boys travel per second on a treadmill is shown to the left. which comparison is accurate? xavier is traveling at 1.5 feet per second. moises is going faster than xavier. the difference in their rates of change is 1 foot per second. if both boys remain on the treadmill for 10 minutes, xavier will have traveled a greater distance.

the distance, in feet, two boys travel per second on a treadmill is shown to the left. which comparison is accurate? xavier is traveling at 1.5 feet per second. moises is going faster than xavier. the difference in their rates of change is 1 foot per second. if both boys remain on the treadmill for 10 minutes, xavier will have traveled a greater distance.

Answer

Explanation:

Step1: Find Xavier's rate

Pick two points on Xavier's line, say (1, 5) and (3, 15). Rate (slope) $m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{15 - 5}{3 - 1}=\frac{10}{2}=5$ feet per second. So Xavier is not traveling at 1.5 feet per second.

Step2: Find Moises's rate

Pick two points on Moises's line, say (1, 2.5) and (3, 7.5). Rate (slope) $m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{7.5 - 2.5}{3 - 1}=\frac{5}{2}=2.5$ feet per second. Since $5>2.5$, Xavier is going faster than Moises.

Step3: Calculate rate - difference

The difference in their rates is $5 - 2.5 = 2.5$ feet per second, not 1 foot per second.

Step4: Calculate distances in 10 minutes

10 minutes = 600 seconds. Xavier's distance $d_X=5\times600 = 3000$ feet. Moises's distance $d_M=2.5\times600=1500$ feet. So if both boys remain on the treadmill for 10 minutes, Xavier will have traveled a greater distance.

Answer:

If both boys remain on the treadmill for 10 minutes, Xavier will have traveled a greater distance.