Question

A baseball diamond is a square 90 ft on a side. a player runs from first base to second at a rate of 16 ​ft/sec. a square has four vertices labeled as follows, given counterclockwise starting at the bottom vertex: home, first base, second base, third base. the line segment between third base and second base is labeled 90 feet. a point is plotted on the line segment between first base and second base labeled player, that is 30 feet from first base, with an arrow drawn from this point in the direction of second base. the distance between the player and second base is labeled x. a line segment labeled y is drawn from third base to the player. the angle between y and x is labeled theta 1. the angle between y and the line segment between third base and second base is labeled theta 2. 30 prime30′ x y x y
a. at what rate is the​ player's distance from third base changing when the player is 3030 ft from first​ base

Answer 1

Answer:

  -8.875 ft/s

Step-by-step explanation:

When the player is 30 ft from first base, they are 60 ft from second base. The angle marked θ1 will have adjacent side 60 ft and opposite side 90 ft, so will satisfy ...

  tan(θ1) = 90/60 = 1.5

  θ1 = arctan(1.5) ≈ 56.3099°

The speed of the player in the direction of third base is the product of the player's speed and the cosine of that angle, so is ...

  y' ≈ (-16 ft/s)cos(56.3099°)

  y' ≈ -8.8752 ft/s . . . . rate of change of player's distance to 3rd base