B2/B3 - 1.3 Homework

The equation X = (+10 m/sec) (t) + 30 m represents the object's position over time. The constant term of 30 m indicates that the object's initial position is 30 m. The coefficient of t is +10 m/sec, which means that the object's position is increasing by 10 m for every second that passes. Therefore, the equation accurately represents the given data and describes the object's position as it moves forward in time.

The given data represents the position of an object at different points in time. To find the position at time 15.0 s, we can use interpolation. By observing the given data, we can see that the position increases by 10 m every 1 s. Therefore, in 15.0 s, the position would have increased by 10 m * 15 = 150 m from the initial position of 30 m. Adding this to the initial position gives us a final position of 30 m + 150 m = 180 m.

When the lines above cross, it means that CMV1 and CMV2 collide or crash into each other.

The object's velocity is constant because the position values increase by a consistent amount of 10 meters every second. Therefore, the object's velocity is +10 m/sec.

not-available-via-ai

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

The runner's position is initially 40 meters at t = 0s. Since the runner is running at a constant pace of 1.5 m/s, we can calculate his position at t = 45s by multiplying the constant pace by the time elapsed. 1.5 m/s * 45s = 67.5 meters. However, this only gives us the distance covered in 45s. To find the runner's position, we need to add this distance to his initial position of 40 meters. 40 meters + 67.5 meters = 107.5 meters. Therefore, the runner's position at t = 45s will be 107.5 meters.

Based on the given data, the object starts at a position of 30 meters. This is evident from the position at time 0.0 seconds, which is given as 30 meters.

The answer is stating that a vehicle cannot be in the same position at the same time value. This is because in physics, an object can only occupy one position at a specific time. Therefore, it is not possible for a vehicle to have multiple positions at the same time value.

At 2-5 seconds, the motion of the vehicle is described as having a velocity of -1 m/sec and traveling in the negative direction. This means that the vehicle is moving backwards at a speed of 1 m/sec.

not-available-via-ai

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

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, so the 5-7 second interval had the greatest velocity.

The interval from 2-5 seconds had the most negative velocity. This can be determined by looking at the graph and observing that the slope of the line during this interval is the steepest and has a negative value. This indicates that the object was moving in the negative direction at a faster rate during this time period compared to the other intervals.

At 10 seconds, the graph shows that the position of the vehicle remains constant. This means that the vehicle is not moving, as its displacement does not change over time. Additionally, since the graph does not show any change in velocity, it can be concluded that the vehicle is at a constant velocity. Therefore, the correct answer is that the vehicle is at a constant velocity and not moving.