*A runner dashes from the starting line (x = 0) to a point 127 m away and then turns around and runs to a point 38 m away from the starting point in 13 seconds. To the nearest tenth of a m/s what is the average speed?
Which is: (127m+89m)/13s=
216m/13s= 16.7m/s
But, the second part of the problem is: What is the runner's average velocity?
Which I'm not sure how to get.
How far to the nearest tenth of a meter can a runner running at 10 m/s run in the time it takes a rock to fall from rest 99 meters?
Which is: (127m+89m)/13s=
216m/13s= 16.7m/s
But, the second part of the problem is: What is the runner's average velocity?
Which I'm not sure how to get.
2 answers
average velocity = (Position at start - position at finish) which is a vector divided by time
In x direction runner goes:
38 meters
so velocity in x direction is 38/13
= 2.92 m/s in x direction
In x direction runner goes:
38 meters
so velocity in x direction is 38/13
= 2.92 m/s in x direction
3 answers