Answer:
[tex]-7\ \text{m/s}[/tex]
Explanation:
x = Distance of 1st car = 8 m
y = Distance of second car = 6 m
[tex]\dfrac{dx}{dt}[/tex] = Velocity of 1st car = 2 m/s
[tex]\dfrac{dy}{dt}[/tex] = Velocity of 2nd car = 9 m/s
Distance between the cars at the given instant
[tex]r=\sqrt{x^2+y^2}\\\Rightarrow r=\sqrt{8^2+6^2}\\\Rightarrow r=10\ \text{m}[/tex]
From pythagoras theorem we have
[tex]r^2=x^2+y^2[/tex]
Differentiating with respect to time we get
[tex]2r\dfrac{dr}{dt}=2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}\\\Rightarrow \dfrac{dr}{dt}=\dfrac{x\dfrac{dx}{dt}+y\dfrac{dy}{dt}}{r}\\\Rightarrow \dfrac{dr}{dt}=\dfrac{-8\times 2-6\times 9}{10}\\\Rightarrow \dfrac{dr}{dt}=-7\ \text{m/s}[/tex]
The rate of change of the distance between the cars at the given instant is [tex]-7\ \text{m/s}[/tex].
