For a better understanding of the solution provided here please find the diagram in the attached file.
In the diagram, A is the intersection. B is the position of the first car and C is the position of the second car. As can be clearly seen, as per the directions given in the question, the cars and the intersection make a right triangle.
The distance between the first car and the intersection is [tex]x[/tex] and the distance between the second car and the intersection is [tex]y[/tex]. The distance between the two cars is depicted by [tex]s[/tex]. As we can see, [tex]s[/tex] is the hypotenuse of the right triangle. At the given instance the distances are 8 meters and 6 meters respectively. Thus, by the Pythagorean Theorem the hypotenuse will be:
[tex]s=\sqrt{8^2+6^2}= \sqrt{100}=10[/tex]...............(Equation 1)
Now, we know that at any instant the Pythagorean Theorem holds and so we will have, in general:
[tex]s^2=x^2+y^2[/tex]
Now, implicitly differentiating the above formula with respect to time, we get:
[tex]2s\frac{ds}{dt}=2x \frac{dx}{dt}+2y \frac{dy}{dt}[/tex]
This can be further simplified by dividing both the sides by the common factor 2 as:
[tex]s\frac{ds}{st}=x \frac{dx}{dy}+y\frac{dy}{dt}[/tex].................(Equation 2)
As we can see from the questions and the diagram, [tex]\frac{dx}{dt}=-2 m/s[/tex] and [tex]\frac{dy}{dt}=-9 m/s[/tex]. The negative sign is there because the distances y and x are reducing as the cars approach the intersection.
Applying this knowledge to (Equation 2) along with the fact that as per (Equation 1), [tex]s=10[/tex], we get (Equation 2) to become:
[tex]10\times \frac{ds}{dt}=8\times -2+6\times -9=-70[/tex]
Therefore, [tex]\frac{ds}{dt}= \frac{-70}{10}=-7 m/s[/tex]
Thus, the rate of change of the distance between the cars at the given instant (in meters per second) is [tex]-7[/tex].
Therefore, out of the given options, option D is the correct one.
