Question

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.

Explanation:

Step1: Calculate Xavier's rate

For Xavier, when $x = 2$, $y = 15$. Rate (speed) $v=\frac{y}{x}$, so $v_{Xavier}=\frac{15}{2}=7.5$ feet per second.

Step2: Calculate Moises's rate

For Moises, when $x = 2$, $y = 5$. Rate (speed) $v=\frac{y}{x}$, so $v_{Moises}=\frac{5}{2} = 2.5$ feet per second.

Step3: Analyze each option

• Option 1: Xavier's rate is 7.5 feet per second, not 1.5 feet per second.
• Option 2: Since $v_{Xavier}=7.5$ feet per second and $v_{Moises}=2.5$ feet per second, Xavier is going faster than Moises.
• Option 3: The difference in their rates of change (speeds) is $7.5 - 2.5=5$ feet per second, not 1 foot per second.
• Option 4: In 10 minutes ($10\times60 = 600$ seconds), distance $d=v\times t$. $d_{Xavier}=7.5\times600 = 4500$ feet and $d_{Moises}=2.5\times600=1500$ feet. So 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.