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.

*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.

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 find the runner's average velocity, you need to divide the total distance covered by the total time taken.

In this case, we can use the formula: average velocity = total distance / total time

You already calculated the total distance covered by the runner, which is 216 meters.

To find the total time taken, we need to figure out how long it took for the rock to fall. We know that the rock fell 99 meters and we need to find the time taken.

We can use the formula: distance = 0.5 * acceleration * time^2

Since the rock was initially at rest, we can consider its initial velocity as 0.

The formula becomes: 99m = 0.5 * 9.8m/s^2 * time^2

Rearranging the formula, we get: time = √(2 * distance / acceleration)

Plugging in the values, we have: time = √(2 * 99m / 9.8m/s^2) ≈ √(20.2041s^2) ≈ 4.5s

Now that we have the total distance covered and the total time taken, we can find the average velocity:

average velocity = 216m / 4.5s ≈ 48m/s

Therefore, the runner's average velocity is approximately 48 meters per second.