Question

Rank in decreasing order the distances traveled by objects having the following pairs of average velocity and time of motion.

I. vavg = +2.0 m/s, t = 2.0 s
II. vavg = +3.0 m/s, t = 2.0 s
III. vavg = -3.0 m/s, t = 3.0 s
A I, II, III
B II, III, I
C II, I, III
D III, II, I

Answer 1

The distances traveled are calculated by multiplying the average velocity by the time of motion. Scenario III has the greatest distance despite the negative velocity, followed by Scenario II and I, making the correct order II, I, III.

Step-by-step explanation:

The question asks us to rank in decreasing order the distances traveled by objects with given average velocities and times. The distance traveled by an object is the product of its average velocity and the time in motion. According to the formula distance = average velocity × time, we can calculate the distances for each scenario as follows:

1. Scenario II: With an average velocity of +3.0 m/s for 2.0 seconds, the distance traveled would be 6.0 meters (3.0 m/s × 2.0 s).
2. Scenario I: With an average velocity of +2.0 m/s for 2.0 seconds, the distance would be 4.0 meters (2.0 m/s × 2.0 s).
3. Scenario III: Although the average velocity here is -3.0 m/s for 3.0 seconds, which indicates motion in the opposite direction, the distance traveled is still 9.0 meters in magnitude (3.0 m/s × 3.0 s).

Since the question asks to rank based on distance traveled and not displacement, the direction indicated by the negative average velocity does not affect the ranking. Hence, the order from greatest to least distance is Scenario III, II, I, which corresponds to the option C II, I, III.