Two cars start moving from the same point.One travels south at mi/h and the other travels west at mi/h.At what rate is the distance between the cars increasing 2 hours later? Round the result to the nearest hundredth.

Question

Two cars start moving from the same point.One travels south at  mi/h and the other travels west at  mi/h.At what rate is the distance between the cars increasing 2 hours later? Round the result to the nearest hundredth.

Quizplus · Accepted Answer

To find the rate of change of distance between two moving objects, we need to use the Pythagorean theorem: $d^2 = x^2 + y^2$, where $d$ is the distance between the two objects, $x$ is the distance traveled by one object, and $y$ is the distance traveled by the other object. To find the rate of change of $d$, we need to differentiate both sides of the equation with respect to time:
$$2d\frac{dd}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}$$
We are given that one car is traveling south at 50 mi/h and the other car is traveling west at 40 mi/h.
Therefore, after 2 hours, the distance traveled by the first car is:
$$x = 50	ext{ mi/h} \cdot 2	ext{ h} = 100	ext{ mi}$$
Similarly, the distance traveled by the second car is:
$$y = 40	ext{ mi/h} \cdot 2	ext{ h} = 80	ext{ mi}$$
The initial distance between the two cars is 0, since they started from the same point.
Plugging these values into our equation and simplifying, we get:
$$\frac{dd}{dt} = \frac{x}{d}\frac{dx}{dt} + \frac{y}{d}\frac{dy}{dt} = \frac{100	ext{ mi}}{\sqrt{100^2 + 80^2}	ext{ mi}}\cdot 0	ext{ mi/h} + \frac{80	ext{ mi}}{\sqrt{100^2 + 80^2}	ext{ mi}}\cdot(-40	ext{ mi/h}) \approx -30.77	ext{ mi/h}$$
Rounding to the nearest hundredth, we get the answer as B.

Two Cars Start Moving from the Same Point