Make a position vs time graph of the following scenario: an object moves at 5.0 m/s for 4.0 s then at -2.0 m/s for 3.0 s.
To make a position vs. time graph for the given scenario, we need to understand how an object's position changes over time.
In this scenario, let's assume that the object starts from an initial position of zero (0) and moves at a constant velocity during each segment.
Segment 1: Object moves at 5.0 m/s for 4.0 seconds.
During this segment, the object moves with a positive velocity (5.0 m/s) for a duration of 4.0 seconds. We can calculate the displacement using the constant velocity equation:
Displacement (Δx) = Velocity (v) × Time (t)
Δx1 = 5.0 m/s × 4.0 s
Δx1 = 20.0 m
This means that during the first segment, the object moves a distance of 20.0 meters.
Segment 2: Object moves at -2.0 m/s for 3.0 seconds.
During this segment, the object moves with a negative velocity (-2.0 m/s) for a duration of 3.0 seconds. Again, we can calculate the displacement:
Δx2 = -2.0 m/s × 3.0 s
Δx2 = -6.0 m
This means that during the second segment, the object moves a distance of -6.0 meters, or it moves 6.0 meters in the opposite direction.
Now, let's plot the position vs. time graph using these calculations:
1. Start at the initial position (0,0).
2. During the first segment (4.0 seconds), the object moves with a constant velocity of 5.0 m/s, resulting in a displacement of 20.0 meters. Draw a straight line with a positive slope from (0,0) to (4.0 s, 20.0 m).
3. During the second segment (3.0 seconds), the object moves with a constant velocity of -2.0 m/s, resulting in a displacement of -6.0 meters. Continue the line from (4.0 s, 20.0 m) with a negative slope to (7.0 s, 14.0 m).
The resulting graph should have two linear segments: one with a positive slope representing the first segment, and another with a negative slope representing the second segment.