Answer
(a)
The rate at which the distance between the vehicles is changing at time $t=0$ is
$\dfrac{100}{\sqrt{13}}\approx27.7350\,km/h$.
(b)
The rate at which the distance between the vehicles is changing at time $t=5\,min$ is
$\dfrac{14300}{\sqrt{16025}}\approx112.9632\,km/h$.
Work Step by Step
(a)
Let the distance between the police car and Sioux Falls be $y$ and the distance between the truck and the Sioux Falls be $x$ at a moment as shown in the given figure.
Now let the distance between the police car and the truck be $z$ at the same moment.
Now use Pythagorean Theorem.
We get, $z^2=x^2+y^2$
Now differentiate with respect to time $t$.
We get, $2z\dfrac{dz}{dt}=2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}$ or $z\dfrac{dz}{dt}=x\dfrac{dx}{dt}+y\dfrac{dy}{dt}$
We have given the speed of the police car and the speed of the truck $160\, km$ and $140\,km$ respectively.
That is, $\dfrac{dy}{dt}=-160 \,km/h$ and $\dfrac{dx}{dt}=140 \,km/h$
Now substitute $\dfrac{dy}{dt}=-160 \,km/h$ and $\dfrac{dx}{dt}=140 \,km/h$ in $z\dfrac{dz}{dt}=x\dfrac{dx}{dt}+y\dfrac{dy}{dt}$.
We get, $z\dfrac{dz}{dt}=x\times 140-y\times 160$
At time $t = 0$, the police car is $20 \,km$ north and the truck is $30\, km$ east of Sioux Falls.
So, $y=20$ and $x=30$
Substitute $y=20$ and $x=30$ in $z^2=x^2+y^2$ and solve for $z$.
We get, $z^2=30^2+20^2=900+400=1300$
$\implies z=\sqrt{1300}=10\sqrt{13}$
Now substitute $z=10\sqrt{13}$, $y=20$ and $x=30$ in $z\dfrac{dz}{dt}=x\times 140-y\times 160$.
We get, $10\sqrt{13}\dfrac{dz}{dt}=30\times 140-20\times 160$
$\implies 10\sqrt{13}\dfrac{dz}{dt}=4200-3200$
$\implies 10\sqrt{13}\dfrac{dz}{dt}=1000$
$\implies \dfrac{dz}{dt}=\dfrac{1000}{10\sqrt{13}}$
$\implies \dfrac{dz}{dt}=\dfrac{100}{\sqrt{13}}\approx27.7350\,km/h$
So, the rate at which the distance between the vehicles is changing at time $t=0$ is
$\dfrac{100}{\sqrt{13}}\approx27.7350\,km/h$.
(b)
Now to calculate $x$ and $y$ at time $t=5 \,min=\dfrac{1}{12} \, hour$.
Use the formula of speed.
That is, $Speed=\dfrac{distance}{time}$
$\implies distance =Speed \times time$
If $Speed=140\,km/h$, then $distance=140\times\dfrac{1}{12}=\dfrac{35}{3}\,km$.
If $Speed=160\,km/h$, then $distance=160\times\dfrac{1}{12}=\dfrac{40}{3}\,km$.
So, We get $x=30+\dfrac{35}{3}\,km=\dfrac{125}{3}\,km$ and $y=20-\dfrac{40}{3}\,km=\dfrac{20}{3}\,km$
Now substitute $x=\dfrac{125}{3}\,km$ and $y=\dfrac{20}{3}\,km$ in $z^2=x^2+y^2$.
We get, $z^2=\left(\dfrac{125}{3}\right)^2+\left(\dfrac{20}{3}\right)^2$
$\implies z^2=\dfrac{15625}{9}+\dfrac{400}{9}$
$\implies z^2=\dfrac{16025}{9}$
$\implies z=\sqrt{\dfrac{16025}{9}}\,km$
Now substitute $z=\sqrt{\dfrac{16025}{9}}\,km$, $x=\dfrac{125}{3}\,km$ and $y=\dfrac{20}{3}\,km$ in $z\dfrac{dz}{dt}=x\times 140-y\times 160$.
We get, $\sqrt{\dfrac{16025}{9}}\dfrac{dz}{dt}=\dfrac{125}{3}\times 140-\dfrac{20}{3}\times 160$
$\implies \sqrt{\dfrac{16025}{9}}\dfrac{dz}{dt}=\dfrac{17500}{3}-\dfrac{3200}{3}$
$\implies \sqrt{\dfrac{16025}{9}}\dfrac{dz}{dt}=\dfrac{14300}{3}$
$\implies \dfrac{\sqrt{16025}}{3}\dfrac{dz}{dt}=\dfrac{14300}{3}$
$\implies \sqrt{16025}\dfrac{dz}{dt}=14300$
$\implies \dfrac{dz}{dt}=\dfrac{14300}{\sqrt{16025}}\approx112.9632\,km/h$
So, the rate at which the distance between the vehicles is changing at time $t=5\,min$ is
$\dfrac{14300}{\sqrt{16025}}\approx112.9632\,km/h$.
