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?

What is the runner's average velocity in the previous problem?

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

To calculate the average speed, we can use the formula:

Average speed = Total distance / Total time

In this case, the runner's total distance is the sum of the distances covered while running away from the starting line and running back towards it. The total distance is 127 m + 38 m = 165 m.

The total time taken is given as 13 seconds.

Plugging these values into the formula, we have:

Average speed = 165 m / 13 s = 12.69 m/s (rounded to the nearest tenth of a m/s)

Now, let's move on to finding the average velocity:

Average velocity is calculated by dividing the displacement (change in position) by the total time taken.

Displacement is the straight-line distance between the starting point and the final position.

In this case, the displacement is given by:

Displacement = Final position - Initial position

The final position is 38 m, and since the runner started at x = 0, the initial position is 0 m.

Therefore, the displacement is 38 m - 0 m = 38 m.

The average velocity is then:

Average velocity = Displacement / Total time

Plugging in the values:

Average velocity = 38 m / 13 s = 2.9 m/s (rounded to the nearest tenth of a m/s)

So, the average speed is 12.7 m/s (rounded), and the average velocity is 2.9 m/s (rounded).