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## The sides of a square baseball diamond are 90 feet long. When a player who is between second and third base is 60 feet from second base and

Question

The sides of a square baseball diamond are 90 feet long. When a player who is between second and third base is 60 feet from second base and heading towards third base at a speed of 22 feet per second, how fast is the distance between the player and home plate changing

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Physics
3 years
2021-08-06T17:30:57+00:00
2021-08-06T17:30:57+00:00 1 Answers
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## Answers ( )

Answer:d(L)/dt = 6,96 ft/s

Explanation:We have a right triangle where the hypotenuse (L) is the distance between the player and home plate, the legs are the line between third base and home plate ( 90 feet ) and distance between the player and third base (x) (over the line between second and third base). So we can write

L² = (90)² + x²

Applying differentiation in relation to time, on both sides of the equation we have:

2*LdL/dt = 0 + 2*x d(x)/dt (2)

In this equation we know:

d(x)/ dt = 22 feet/sec

x = 30 ft

We need to calculate L when the player is at 30 feet from third base

Then

L² = (90)² + (30)²

L² = 8100 + 900

L = √9000

L =94,87 feet

Then we are in condition for calculate d(L)/dt from the equation

2*Ld(L)/dt = 0 + 2*x d(x)/dt

2*94,87 * d(L)/dt = 2* 30* 22 ⇒ 189,74 d(L)/dt = 1320

d(L)/dt = 1320/ 189,74

d(L)/dt = 6,96 ft/s