I am given a velocity versus time graph for a runner. For it, I had to figure out the average velocity for the entire 16 s interval (for this, I got 2 m/s). Now they want me to find the displacement of the runner in the first 15.4 s.
I know that it must be somewhere around 30 (since 2*15=30), but how can I factor in the .4 of a second?
Isn't the area underneath the velocity plot distance? Velocity*time= distance?
I think this is what they are wanting you to do, measure the area underneath. An easy way to do that is to count the squares on the graph paper.
Great I got it, thanks. :)
To find the displacement of the runner in the first 15.4 seconds, you can use the average velocity you calculated for the entire 16-second interval as an approximation. Since you found that the average velocity is 2 m/s, you can assume that the runner's velocity remained constant at 2 m/s during the 15.4-second interval.
To calculate the displacement, you can multiply the average velocity by the time interval. In this case, you would multiply 2 m/s by 15.4 seconds:
Displacement = Average Velocity × Time Interval
= 2 m/s × 15.4 s
= 30.8 meters
Therefore, while the approximation based on the average velocity suggests a displacement of around 30 meters, including the additional 0.4 seconds gives a more precise result of 30.8 meters.