A police car is driving northward, chasing a vehicle which has just taken a right turn at an intersection, and is travelling due west. When the police car is 600 yards south of the intersection, and the car is 800 yards east of the intersection, an aerial speed radar indicates that the distance between the two cars is increasing by 20 mph. If the police car is travelling at 60 mph when the measurment is taken, how fast is the other car going?
To solve this problem, we can use the concept of relative velocity. The relative velocity is the velocity of one object with respect to another object.
Here's how we can solve the problem step by step:
Step 1: Determine the relative velocity of the police car with respect to the other vehicle.
The relative velocity is the difference between the velocities of the police car and the other vehicle. Since the police car is traveling northward at 60 mph and the distance between the two cars is increasing by 20 mph, the relative velocity is 60 mph + 20 mph = 80 mph. This means the other car is moving away from the police car at a speed of 80 mph.
Step 2: Break down the relative velocity into components.
Since the police car is traveling northward and the other car is traveling westward, we need to break down the relative velocity into its northward and westward components.
Let's say the northward component is Vn and the westward component is Vw.
Step 3: Determine the distances traveled by both cars.
From the information given, we know that when the police car is 600 yards south of the intersection, and the other car is 800 yards east of the intersection, the distance between the two cars is increasing by 20 mph.
The distance that the police car travels southward, Sp, is 600 yards, while the distance that the other car travels eastward, Sw, is 800 yards.
Step 4: Use the relative velocity components and distances traveled to find the speed of the other car.
Now we can use the relative velocity components and distances traveled to get the speed of the other car.
Using the concept of relative velocity, we have the following equations:
Vn = 80 mph
Sp = 600 yards
Vw = ?
We can use the Pythagorean theorem to relate the distances traveled by both cars:
Sp^2 + Sw^2 = (600 yards)^2 + (800 yards)^2
From the Pythagorean theorem, we have:
Sp^2 + Sw^2 = (600)^2 + (800)^2
Sp^2 + Sw^2 = 360000 + 640000
Sp^2 + Sw^2 = 1000000
Since the relative velocity is perpendicular to both distances traveled, the distances traveled are perpendicular to each other, which allows the use of the Pythagorean theorem.
By substituting the known values, we get:
600^2 + Sw^2 = 1000000
Sw^2 = 1000000 - 600^2
Sw^2 = 1000000 - 360000
Sw^2 = 640000
Sw = √640000
Sw = 800 yards
Since the relative velocity is the rate of change of the distance between the two cars, we have:
Relative velocity V = sqrt(Vn^2 + Vw^2)
Substituting the values, we get:
80 = sqrt(60^2 + Vw^2)
80 = sqrt(3600 + Vw^2)
80^2 = 3600 + Vw^2
6400 = 3600 + Vw^2
Vw^2 = 6400 - 3600
Vw^2 = 2800
Vw = √2800
Vw = 52.92 mph (approximately)
Therefore, the speed of the other car is approximately 52.92 mph.