Two cars leave a town at the same time and travel at constant speeds along straight roads that meet at an angle of 60° in the town. If one car travels twice as fast as the other and the distance between them increases at the rate of 63 mi/h, how fast is the slower car traveling?

36 mph

That's what I get, too.

Can I please see your solution?Thanks!

36

To solve this problem, we can break it down into two parts: determining the rate at which the distance between the cars is changing, and then finding the speed of the slower car.

Let's start by finding the rate at which the distance between the cars is changing. We are given that this rate is 63 mi/h.

Now, let's consider the speeds of the two cars. Let's say the slower car is traveling at a speed of s mi/h. Since the faster car is traveling at twice the speed of the slower car, its speed is 2s mi/h.

Next, let's focus on the motion of the slower car. Since angle in the town where the roads meet is 60°, we can use trigonometry to relate the distance between the cars and the distance traveled by the slower car.

Let D be the distance between the two cars at a specific time, and let x be the distance traveled by the slower car at the same time. From trigonometry, we can write:

D = x.cos(60°)

Now, we can differentiate both sides of the equation with respect to time. Let t be the time:

dD/dt = dx/dt.cos(60°)

We know that dD/dt = 63 mi/h and cos(60°) = 0.5. Therefore, we have:

63 = dx/dt * 0.5

Rearranging the equation, we find:

dx/dt = 63 / 0.5

dx/dt = 126 mi/h

So, the rate at which the slower car is traveling is 126 mi/h.

Therefore, the slower car is traveling at a speed of 126 mi/h.