Answer:
13.86 ft/sec
Step-by-step explanation:
If we let x = distance batter has run at time t and D = distance from second base to the batter at time t, then we know [tex]\frac{dx}{dt}=31 ft/s[/tex] and we want [tex]\frac{dD}{dt}[/tex] when he is halfway (at x = 45).
Using Pythagoras theorem
[tex]D^2=90^2+(90-x)^2\\\text{Differentiating with respect to t}\\2D\frac{dD}{dt}=0+2(90-x)(-1)\frac{dx}{dt}\\2D\frac{dD}{dt}=-2(90-x)\frac{dx}{dt}[/tex]
When x=45
[tex]D^2=90^2+(90-x)^2\\D^2=90^2+(90-45)^2\\D^2=10125\\D=\sqrt{10125}[/tex]
[tex]2D\frac{dD}{dt}=-2(90-x)\frac{dx}{dt}\\2\sqrt{10125}\frac{dD}{dt}=-2(90-45)(31)\\\frac{dD}{dt}=\frac{-2(45)(31)}{2\sqrt{10125}} =-13.86ft/s[/tex]
Thus, when the batter is halfway to first base, the distance between second base and the batter is decreasing at the rate of about 13.86 ft/sec.
