as Sue runs, her knee joint passes through the following vertical positions (meters):
Frame 1: .043
Frame 2: .145
Frame 3: .21
Frame 4: .39
Frame 5: .415
The data were collected at a frequency of 100 Hz. What was the vertical velocity of her knee joint at the 2nd time point?
from t = 0 to t = .02 she goes(.21 - .043) = .167
.167 m / .02 seconds = 8.35 m/s
To determine the vertical velocity of Sue's knee joint at the 2nd time point, we need to calculate the change in position (displacement) divided by the change in time.
First, let's identify the frame numbers and corresponding positions:
Frame 1: .043 meters
Frame 2: .145 meters
Frame 3: .21 meters
Frame 4: .39 meters
Frame 5: .415 meters
Next, we need to determine the change in position. Since the data was collected at a frequency of 100 Hz, the time interval between each frame is 1/100 seconds.
The change in position between frame 1 and frame 2 is:
.145 - .043 = .102 meters
To calculate the change in time, we multiply the time interval (1/100 seconds) by the frame number difference:
1/100 seconds * (2 - 1 frames) = 1/100 seconds
Now we can calculate the vertical velocity:
Vertical velocity = Change in position / Change in time
Vertical velocity = .102 meters / (1/100 seconds)
Vertical velocity = .102 meters * (100/1 seconds)
Vertical velocity = 10.2 m/s
Therefore, the vertical velocity of Sue's knee joint at the 2nd time point is 10.2 m/s.