Question

a baseball diamond is a square with side 90 feet. if a batter hits the ball and runs towards first base with a speed of 25 ft/sec, at what speed is his distance from second base decreasing when he is two thirds of the way to first base?

Answer 1

The rate when his distance from second base changes and it is two third of the way to first base is - 13.86 feet/s.

let us consider the following values

distance between batter to Second base = d feet

distance between batter to First base

= x feet

distance between first to second base

= 90 feet

Using the Pythagorean theorem:

d² = x² + 90²

Knowing that the velocity of the bat to first base is 25 ft/s,

the distance x can be written as: x

= 90 - 25t

The equation for is,d² = 90²+(90 -25t)²--(1)

we want to find the rate of change of d.

let it be d' and from implicit differentiation:

2d.d' = 2(90-24t) × (-25)

d' = -50(90 - 25t) / (2d)

d' = -25 (90 - 24t) / d ---(*)

We can determine t and d from the fact that the batter is two third to his first base: d = √90^2 + 60^2 = √1 = 108.2 feet

putting the values of d in (1) we get,

=> 90 - 25t = 60 => 25t = 30 => t = 6/5.

Substitute d' in the formula equation (*)

d' = -25 (90 - 24t) / d = -25(90-30)/108.2

d'= - 13.86

d' = - 13.86 feet/s

negative sign means decreasing distance from second base (as it approaches the first

bases, it approaches the second base).

Hence,The required speed is 13.86 feet/s

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