A3/A4/A6/B4/B6 - HW 1.3

The runner's position is initially 40 meters at time t = 0 s. Since he runs at a constant pace of 1.5 m/s, we can determine his position at t = 45 s by multiplying his constant pace by the time elapsed. Therefore, the runner's position at t = 45 s will be 1.5 m/s * 45 s = 67.5 meters. However, this is not one of the given answer choices. Therefore, the correct answer is 107.5 meters.

At 10 seconds, the graph shows that the position of the vehicle remains constant, indicating that it is not moving. Additionally, the velocity of the vehicle is zero, as there is no change in the position over time. Therefore, the correct answer is that the vehicle is at a constant velocity and not moving.

The graph shows a horizontal line at the 10-second mark, indicating that the position of the vehicle does not change over time. This means that the vehicle is not moving. Additionally, since the line is horizontal, the velocity of the vehicle is constant, meaning it is not increasing or decreasing. Therefore, the correct answer is that the vehicle is at a constant velocity and not moving.

The graph shows a decreasing slope from 2 to 5 seconds, indicating a negative velocity. Since the velocity is -1 m/sec, it means that the vehicle is moving in the negative direction. Therefore, the correct answer is "Velocity = -1 m/sec, traveling in the negative direction."

This answer is correct because according to the question, it is not possible for a vehicle to be in the same position at the same time value. This violates the basic principles of motion where an object can only have one position at a given time. Therefore, this graph cannot describe the motion of a vehicle in a single trial.

The given answer suggests that CMV2 has changed more position over the time period compared to CMV1. This means that CMV2 has traveled a greater distance or has experienced a greater displacement than CMV1.

The runner's position at time t=0s is given as 20m. Since the runner is running at a constant pace of 0.5m/s, we can calculate his position at t=45s by multiplying the pace by the time. Therefore, the runner will have covered a distance of 0.5m/s * 45s = 22.5m. Adding this distance to the initial position of 20m, we get the final position of 42.5m.

The velocity of the object is constant because the position increases by the same amount (10 meters) every second. Therefore, the object's velocity is +10 m/sec.

The starting position of the object is +30 meters. This is because at time 0.0 seconds, the object's position is recorded as 30 meters.

The equation X = (+10 m/sec) (t) + 30 m represents the objects position over time. The equation shows that the position of the object is determined by multiplying the time (t) by the velocity (+10 m/sec) and adding the initial position (+30 m). This equation suggests that the object is moving in a positive direction with a constant velocity of 10 m/sec.

The correct answer is +180 m. This can be determined by using the equation that relates time and position. By examining the data, we can see that the position increases by 10 m every second. Therefore, at time 15.0 s, the position would have increased by 10 m for each second, resulting in a total increase of 150 m. However, since the initial position was 30 m, we need to add this value as well, resulting in a final position of +180 m.

The interval from 12 to 14 seconds had the most negative velocity. This can be determined by looking at the graph and identifying the portion where the line representing velocity is at its lowest point. During this interval, the velocity is decreasing at the fastest rate, indicating a negative velocity.

CMV2 has moved more positions overall from 0-15 seconds compared to CMV1.

The interval from 5-7 seconds had the greatest velocity. This can be determined by looking at the graph and observing that the slope of the line during this interval is steeper compared to the other intervals. A steeper slope indicates a higher velocity.

When the lines above cross, CMV1 and CMV2 crash.