Question

the distance traveled during kevins trip is shown in the line graph.

during which leg of the trip was the average speed the fastest?

a. leg 1

b. leg 2

c. leg 3

d. leg 4

Explanation:

Step1: Recall Speed Formula

Average speed is calculated as \( \text{speed} = \frac{\text{distance}}{\text{time}} \). On a distance - time graph, the slope of the line segment representing a leg of the trip is equal to the average speed (since slope \(=\frac{\Delta y}{\Delta x}=\frac{\text{change in distance}}{\text{change in time}}\)). So, we need to find the leg with the steepest slope.

Step2: Analyze Each Leg's Slope

• Leg 1: From \( t = 0 \) to \( t = 4 \) hours. Let's assume the distance at \( t = 0 \) is \( 0 \) and at \( t = 4 \) is \( 5 \) (from the graph). The change in time \( \Delta t=4 - 0 = 4 \) hours, change in distance \( \Delta d = 5-0 = 5 \) km. Slope (speed) \(=\frac{5}{4}=1.25\) km/h.
• Leg 2: From \( t = 4 \) to \( t = 6 \) hours. Distance at \( t = 4 \) is \( 5 \), at \( t = 6 \) is \( 15 \). \( \Delta t=6 - 4 = 2 \) hours, \( \Delta d=15 - 5 = 10 \) km. Slope (speed) \(=\frac{10}{2}=5\) km/h.
• Leg 3: From \( t = 6 \) to \( t = 14 \) hours. Distance at \( t = 6 \) is \( 15 \), at \( t = 14 \) is \( 20 \). \( \Delta t=14 - 6 = 8 \) hours, \( \Delta d=20 - 15 = 5 \) km. Slope (speed) \(=\frac{5}{8}=0.625\) km/h.
• Leg 4: From \( t = 14 \) to \( t = 18 \) hours. Distance at \( t = 14 \) is \( 20 \), at \( t = 18 \) is \( 30 \). \( \Delta t=18 - 14 = 4 \) hours, \( \Delta d=30 - 20 = 10 \) km. Slope (speed) \(=\frac{10}{4}=2.5\) km/h.

Comparing the slopes (speeds): \( 5>2.5 > 1.25>0.625 \). So Leg 2 has the steepest slope and thus the fastest average speed.

Answer:

B. Leg 2