Answer
$ \frac{520-8\sqrt {65}}{65} \times 0.1415977773=0.9922778766 mi/h$
Work Step by Step
Changes occur along the direction of East to the West and at an altitude of $\frac{1}{2}$ miles
in the direction of South to North.
Refrence to the fig
$EC>2$ The Car is more than 2 miles away.
$LC=2$ The Car is exactly at 2 miles
$FC<2$ Distance of the Car is less than 2 miles
So
$EC=x+2$
$LC=2$
$FC=x-2$
$(AC)^2=(AE)^2+(EC)^2=(\frac{1}{2})^2+(x+2)^2=\frac{1}{4}+x^2+4x+4=x^2+4x+ \frac{17}{4}$
$AC=\sqrt {x^2+4x+\frac{17}{4}}$
$(BC)^2=(BF)^2+(FC)^2=(\frac{1}{2})^2+(x-2)^2=\frac{1}{4}+x^2-4x+4=x^2-4x+\frac{17}{4}$
$BC=\sqrt {x^2-4x+\frac{17}{4}}$
Reference to the fig
$(AB)^2=(AC)^2+(BC) ^2-2 (AC)(BC)\cos\theta$ ............................ eq (1)
Putting the values of AC,, BC in equation (1)
$(AB)^2=(\sqrt {x^2+4x+\frac{17}{4}})^2+( \sqrt {x^2-4x+\frac{17}{4}})^2 - 2\times\sqrt {x^2+4x+\frac{17}{4}} \times \sqrt {x^2-4x+\frac{17}{4}} \cos\theta$
Putting $AB=y$
$y^2=x^2+4x+\frac{17}{4}+x^2-4x+\frac{17}{4}- 2\times\sqrt {x^2+4x+\frac{17}{4}} \times \sqrt {x^2-4x+\frac{17}{4}}$
$y^2=2x^2+\frac{17}{2}- 2\times\sqrt {x^2+4x+\frac{17}{4}} \times \sqrt {x^2-4x+\frac{17}{4}}\cos\theta$
$y^2=2x^2+\frac{17}{2}- 2 \sqrt {([x^2+\frac{17}{4})+4x] [(x^2+\frac{17}{4})-4x] }\cos\theta$
$y^2= 2x^2+\frac{17}{2} -2 \sqrt { [x^2+\frac{17}{4}]^2-16x^2 }\cos\theta$
$y^2= 2x^2+\frac{17}{2} -2 \sqrt { x^4+\frac{17}{2}x^2-16x^2+\frac{289}{16} } \cos\theta$
$y^2= 2x^2+\frac{17}{2} -2 \sqrt { x^4-\frac{15}{2}x^2+\frac{289}{16} } \cos\theta$ ............... eq(2)
Putting $x=2$ in equation (2)
$y^2=\frac{33}{2}- 2 \sqrt { \frac{65}{16}}\cos\theta$ Or
$y=\sqrt { \frac{33}{2}- 2 \sqrt { \frac{65}{16}}\cos\theta }$
Taking derivative with respect to t of equation (2)
$2y\frac{dy}{dt}=4x-2( \frac{1}{2})[x^4-\frac{15}{2}x^2+\frac{289}{16}]^{-\frac{1}{2}}(4x^3-15x)\cos\theta - 2 \sqrt { x^4-\frac{15}{2}x^2+\frac{289}{16} }(-\sin\theta)\frac{d\theta}{dt}$
$2y\frac{dy}{dt}=4x-[x^4-\frac{15}{2}x^2+\frac{289}{16}]^{-\frac{1}{2}}(4x^3-15x)\cos\theta +2 \sqrt { x^4-\frac{15}{2}x^2+\frac{289}{16} }(\sin\theta)\frac{d\theta}{dt}$ ............. eq (3)
Putting $y=\sqrt { \frac{33}{2}- 2 \sqrt { \frac{65}{16}}\cos\theta }$, $x=2$ in equation (3)
$2 \sqrt { \frac{33}{2}- 2 \sqrt { \frac{65}{16}}\cos\theta } \frac{dy}{dt}=8-[16-30+\frac{289}{16}]^{-\frac{1}{2}}(32-30)\cos\theta +2 \sqrt { 16-30+\frac{289}{16} }(\sin\theta)\frac{d\theta}{dt}$
When $t\longrightarrow$0, $\theta \longrightarrow$ 0 $\Longrightarrow$
Also $\cos0=1$, $\sin0=0$
$2 \sqrt { \frac{33}{2}- 2 \sqrt { \frac{65}{16}} } \frac{dy}{dt}=8-[16-30+\frac{289}{16}]^{-\frac{1}{2}}(32-30)$
On simplification
$ \frac{dy}{dt}=\frac{520-8\sqrt {65}}{65} \times 0.1415977773=0.9922778766 mi/h$
