Question #329755

A baseball player is running from the second base to the third base at 20ft/sec. At what rate is his

distance from the home plate changing when he is 30ft from the third base. The baseball diamond

is a square 90ft on aside.


1
Expert's answer
2022-04-17T15:13:22-0400

The picture represents a baseball diamond and a player.



The player is located at the point EE. He is between the third (C)(C) and the second base (BB ). The arrow shows the velocity vector. CE=30ft|CE|=30ft. AE|AE| is the distance between the home plate and the player. Using the Pythagorean theorem, we get: AE2=AB2+EB2=902+602=11700|AE|^2=|AB|^2+|EB|^2=90^2+60^2={11700}. AF|AF| denotes the distance between the home plate and the player, when he will pass a small distance EF|EF|. EF=vΔt|EF|=v\Delta t, where Δt\Delta t is a small period of time. AF2=AB2+(FE+BE)2=902+(vΔt+60)2|AF|^2=|AB|^2+(|FE|+|BE|)^2=90^2+(v\Delta t+60)^2. It is enough to find the derivative of AFAE|AF|-|AE| with respect to Δt\Delta t and substitute Δt=0.\Delta t=0. It will show how fast does the distance change. AFAE=AF=1AFv(vΔt+60)|AF|'-|AE|'=|AF|'=\frac{1}{|AF|}v(v\Delta t+60). After setting Δt=0\Delta t =0 we get: 60v1AE=111700602011.0960v\frac{1}{|AE|}=\frac{1}{\sqrt{11700}}60\cdot20\approx11.09. Thus, the distance between the home plate and the player changes with the velocity 11.09ft/sec11.09ft/sec. This is the velocity, when the player is at the point EE . It changes with time.


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