A person on horseback moves according to the velocity-versus-time graph shown in Figure 2-32. (The vertical axis is marked in increments of 1 m/s and the horizontal axis is marked in increments of 7 s.) Find the displacement of the person for each of the segments A, B, and C.
sA =
sB =
sC =
A has to coordinates Velocity 1m/s time 14s
B has the coordinates V = 3m/s
time 21s
C has the coordinates V = 1m/s
time 35s
To find the displacement for each segment A, B, and C, we need to calculate the area under the velocity-time graph.
Segment A:
The coordinates for segment A are velocity = 1 m/s and time = 14 s. Since the velocity is constant at 1 m/s, segment A represents a rectangular area. The width of the rectangle is the time duration, which is 14 s. The height of the rectangle is the velocity, which is 1 m/s. Therefore, the area of segment A is:
Area(A) = velocity * time = 1 m/s * 14 s = 14 meters
So, the displacement for segment A is 14 meters.
Segment B:
The coordinates for segment B are velocity = 3 m/s and time = 21 s. Similar to segment A, segment B also represents a rectangular area. The width of the rectangle is the time duration, which is 21 s, and the height is the velocity, which is 3 m/s. The area of segment B is:
Area(B) = velocity * time = 3 m/s * 21 s = 63 meters
Therefore, the displacement for segment B is 63 meters.
Segment C:
The coordinates for segment C are velocity = 1 m/s and time = 35 s. Again, segment C represents a rectangular area with a width of 35 s and a height of 1 m/s. The area of segment C is:
Area(C) = velocity * time = 1 m/s * 35 s = 35 meters
Thus, the displacement for segment C is 35 meters.
Therefore, the displacement for segment A is 14 meters, segment B is 63 meters, and segment C is 35 meters.
To find the displacement of the person for each segment (A, B, and C), we need to calculate the area under the velocity-versus-time graph for each segment.
For Segment A:
The coordinates given are:
Velocity = 1 m/s
Time = 14 s
To find the displacement (sA), we need to calculate the area under the graph for segment A. Since the graph is a straight line, the area will be the rectangle formed by the velocity and time.
Area = Velocity x Time = 1 m/s x 14 s = 14 meters
Thus, sA = 14 meters.
For Segment B:
The coordinates given are:
Velocity = 3 m/s
Time = 21 s
Again, we need to calculate the area under the graph for segment B. Since the graph is a straight line, the area will be the rectangle formed by the velocity and time.
Area = Velocity x Time = 3 m/s x 21 s = 63 meters
Thus, sB = 63 meters.
For Segment C:
The coordinates given are:
Velocity = 1 m/s
Time = 35 s
Similarly, we calculate the area under the graph for segment C.
Area = Velocity x Time = 1 m/s x 35 s = 35 meters
Thus, sC = 35 meters.
Therefore, the displacement for segment A (sA) is 14 meters, for segment B (sB) is 63 meters, and for segment C (sC) is 35 meters.
It is really pretty simple: Displacement is the area under the velocity-time graph
(remember in the 5th grade...distance=velocity*time?)
So count the squares under the graph, each one is (1m/s*7sec=7m)