Two cars are travelling east along a straight highway at the same speed. At an intersection, the highway branches and splits into two straight roads, one heading approximately north-east and one heading approximately south-east. The angle betwen the two roads is 60 degrees. The two cars leave the intersection at the same time, one travelling north-east at 100 km/h and the other travelling south-east at 80 km/h. How fast is the distance between the two cars changing 30 minutes after they leave the intersection.
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To solve this problem, we can use the concept of related rates. We are given that the two cars are traveling at different speeds and we need to find how fast the distance between them is changing.
Let's denote the distance between the two cars as "d". We are asked to find the rate at which d is changing.
To solve this problem, we can first find expressions for how the positions of the two cars change with respect to time and then differentiate these expressions to determine how d is changing.
Let's assume that 30 minutes after leaving the intersection, the two cars have traveled distances x and y along their respective paths.
Since car A is traveling at 100 km/h for 30 minutes, it will have traveled a distance of (100/60) * 30 = 50 km.
Similarly, car B is traveling at 80 km/h for 30 minutes, so it will have traveled a distance of (80/60) * 30 = 40 km.
Now we have two right-angled triangles formed by the two cars and the distance between them. Let's consider the triangle formed by car A, car B, and the point where the roads split.
In this triangle, the distance between the two cars forms the hypotenuse (d), and the distances traveled by car A and car B are the legs (x and y, respectively).
We can use the Pythagorean theorem to relate d, x, and y in this triangle:
d^2 = x^2 + y^2 (Equation 1)
Differentiating both sides of Equation 1 with respect to time t, we get:
2d (dd/dt) = 2x (dx/dt) + 2y (dy/dt)
Simplifying this equation, we get:
d (dd/dt) = x (dx/dt) + y (dy/dt) (Equation 2)
Since the two cars are traveling at constant speeds and in opposite directions, dx/dt = -100 km/h and dy/dt = 80 km/h. Substituting these values into Equation 2, we have:
d (dd/dt) = x (-100 km/h) + y (80 km/h)
Now, let's substitute the values of x and y that we calculated earlier:
d (dd/dt) = 50 km (-100 km/h) + 40 km (80 km/h)
Simplifying further:
d (dd/dt) = -5000 km^2/h + 3200 km^2/h
d (dd/dt) = -1800 km^2/h
Finally, we can solve for dd/dt by dividing both sides of the equation by d:
dd/dt = (-1800 km^2/h) / d
Substituting the value of d that we found earlier:
dd/dt = (-1800 km^2/h) / √(x^2 + y^2)
After substituting the values of x and y, we can evaluate dd/dt. However, we don't have enough information to find the exact value of d, so we cannot calculate the exact rate at which the distance between the two cars is changing.
To find the rate at which the distance between the two cars is changing, we can first visualize the problem.
Let's assume that the intersection is located at the origin of a coordinate system, with the north-east road being the positive x-axis, and the south-east road being the positive y-axis. The two cars start at the intersection and travel along their respective roads.
After 30 minutes, the car traveling north-east would have covered a distance of (100 km/h) * (0.5 h) = 50 km. Similarly, the car traveling south-east would have covered a distance of (80 km/h) * (0.5 h) = 40 km.
We can represent the positions of the cars at time t as (x, y), where x is the distance traveled by the north-east car and y is the distance traveled by the south-east car.
At any given time t, the position of the north-east car is (100t, 0) (since it only travels along the x-axis), and the position of the south-east car is (40t, -40t) (since it travels along the y-axis with a negative slope).
The distance d between the two cars at time t is given by the distance formula:
d = sqrt((100t - 40t)^2 + (0 - (-40t))^2)
= sqrt(60^2 + 40^2)t
= sqrt(3600 + 1600)t
= sqrt(5200)t
= 2sqrt(1300)t
To find the rate at which the distance between the two cars is changing, we differentiate the equation with respect to time t:
dd/dt = 2sqrt(1300)
Therefore, the distance between the two cars is changing at a constant rate of 2sqrt(1300) km/h.