Runners stand at first and second base in a baseball game. at the moment a ball is hit, the runner at first base runs to second base at 18 ft/sec. simultaneously the runner on second base runs to third base at 20 ft/sec. how fast is the distance between the runners changing 1 sec after the ball is all steps.

In one second the distance between the player at first base and the player at second base is extended by 2 feet.

Since runners stand at first and second base in a baseball game, and at the moment a ball is hit, the runner at first base runs to second base at 18 ft / sec, and simultaneously the runner on second base runs to third base at 20 ft / sec, to determine how fast is the distance between the runners changing 1 sec after the ball is hit, the following calculation must be performed:

Therefore, in one second the distance between the player at first base and the player at second base is extended by 2 feet.

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decreasing at 11.99 ft/s

Step-by-step explanation:

The path from 1st base to 2nd base and that from 2nd base to 3rd base form the legs of a right triangle. The hypotenuse of that triangle is the distance between the runners.

If the distance between bases is 90 ft (as in major-league baseball), then the runner on first has distance from second of ...

... 90 -18t . . . . . . . where t is the time in seconds since he left 1st base

The runner on second has a distance from second of

... 20t . . . . . . . . . . where t is the time in seconds since he left 2nd base

Then the Pythagorean theorem tells us the distance between the runners is

... d = √((90 -18t)² +(20t)²)

The rate of change of distance between the runners is the derivative of this with respect to time.

The distance between runners is decreasing at 11.99 ft/s one second after they leave the bases.

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Comment on the solution

If you look carefully at the derivative expression, you find it is the sum of each of the speeds multiplied by the cosine of the angle between the direction of running and the direction of the other runner. As we found above, the distance between runners is √5584 ≈ 74.7262 ft, so the runner from first base is decreasing the distance at the rate of ...

... (18 ft/s)×(72 ft)/(74.7262 ft) ≈ 17.3433 ft/s

At the same time, the runner from second base is increasing the distance between them at ...

... (20 ft/s)×(20 ft)/(74.7262 ft) ≈ 5.3529 ft/s

So, the distance between runners is changing at the rate of

... -17.3433 +5.3529 = -11.9904 . . . . ft/s

No derivatives are necessary in this approach to the solution.