Car A leaves an intersection at the same time as Car B. Car A is heading west at 50km/h and Car B is heading north at 70km/h. Determine the rate of change of the distance between the two cars after 20min
The "geometry" of this problem is a simple right-angled triangle.
At a time of t hours,
let the distance travelled north be 70t km
let the distance travelled west be 50t km
let the distance between them be d km
clearly,
d^2 = (70t)^2 + (50t)^2 = 7400t^2
d = √7400 t
dd/dt = √7400 <----- a constant
= 86.02 km/h
As long as they maintain the same speed in their same direction
they will separate at a constant rate of 86.02 km/h
To determine the rate of change of the distance between the two cars after 20 minutes, we need to calculate the rate at which the distance between the two cars is changing.
First, let's convert 20 minutes to hours. There are 60 minutes in an hour, so 20 minutes is equal to 20/60 = 1/3 hour.
Car A is moving west at a speed of 50 km/h. This means that after 1/3 hour, Car A will have traveled a distance of 50 * (1/3) = 50/3 km.
Car B is moving north at a speed of 70 km/h. Similarly, after 1/3 hour, Car B will have traveled a distance of 70 * (1/3) = 70/3 km.
Now, to find the distance between the two cars, we can use the Pythagorean theorem. The distance d between the two cars can be found using the formula:
d = sqrt((distance traveled by Car A)^2 + (distance traveled by Car B)^2)
Substituting the calculated values:
d = sqrt((50/3)^2 + (70/3)^2)
Now we can solve this equation to find the distance between the two cars after 20 minutes.
To determine the rate of change of the distance between the two cars after 20 minutes, we need to convert the time to hours.
20 minutes is equal to 20/60 = 1/3 hour.
Let's assume the initial distance between the two cars is 0.
After 1/3 hour, Car A will have traveled a distance of 50 km/h * 1/3 h = 50/3 km west.
After 1/3 hour, Car B will have traveled a distance of 70 km/h * 1/3 h = 70/3 km north.
Now, we can use the Pythagorean theorem to find the distance between the two cars after 1/3 hour:
distance^2 = (50/3)^2 + (70/3)^2
distance^2 = 2500/9 + 4900/9
distance^2 = 7400/9
distance = sqrt(7400/9)
distance ≈ 30.55 km
So, after 20 minutes, the distance between the two cars is approximately 30.55 km.
To find the rate of change of the distance, we need to find the derivative of the distance function with respect to time:
d(distance) / dt = (d/dt) sqrt(7400/9)
Using the chain rule, we can simplify this expression:
d(distance) / dt = (1/2) * (7400/9)^(-1/2) * (d/dt) (7400/9)
d(distance) / dt = (1/2) * (7400/9)^(-1/2) * 0
d(distance) / dt = 0
Therefore, the rate of change of the distance between the two cars after 20 minutes is 0 km/h.