Question

Two cars are driving towards an intersection from perpendicular directions.

Answer 1

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 $x$ and the distance between the second car and the intersection is $y$. The distance between the two cars is depicted by $s$. As we can see, $s$ 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:

$s=\sqrt{8^2+6^2}= \sqrt{100}=10$...............(Equation 1)

Now, we know that at any instant the Pythagorean Theorem holds and so we will have, in general:

$s^2=x^2+y^2$

Now, implicitly differentiating the above formula with respect to time, we get:

$2s\frac{ds}{dt}=2x \frac{dx}{dt}+2y \frac{dy}{dt}$

This can be further simplified by dividing both the sides by the common factor 2 as:

$s\frac{ds}{st}=x \frac{dx}{dy}+y\frac{dy}{dt}$.................(Equation 2)

As we can see from the questions and the diagram, $\frac{dx}{dt}=-2 m/s$ and $\frac{dy}{dt}=-9 m/s$. 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), $s=10$, we get (Equation 2) to become:

$10\times \frac{ds}{dt}=8\times -2+6\times -9=-70$

Therefore, $\frac{ds}{dt}= \frac{-70}{10}=-7 m/s$

Thus,  the rate of change of the distance between the cars at the given instant (in meters per second) is $-7$.

Therefore, out of the given options, option D is the correct one.