Two cars start moving from the same point. One travels south at 28mi/h and the other travels west at 50mi/h. At what rate is the distance between the cars increasing 4 hours later?
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To find the rate at which the distance between the cars is increasing, we can use the concept of related rates.
Let's assume that the two cars start moving at time t = 0 and the distance between them at that time is d. We are asked to find the rate at which the distance between the cars is increasing, which we can represent as dd/dt (the derivative of the distance with respect to time).
Since one car is traveling south and the other is traveling west, we can visualize their respective motions as forming a right triangle. The distance between the cars is the hypotenuse of this right triangle.
We can use the Pythagorean theorem to relate the distance between the cars (d) to the distances traveled by each car. According to the Pythagorean theorem:
d^2 = (distance traveled by the southbound car)^2 + (distance traveled by the westbound car)^2
Let's consider the distance traveled by each car after 4 hours. The southbound car is traveling at a constant speed of 28 miles per hour, so it will have covered a distance of (28 miles/hour) * (4 hours) = 112 miles.
Similarly, the westbound car is traveling at a constant speed of 50 miles per hour, so its distance traveled will be (50 miles/hour) * (4 hours) = 200 miles.
Plugging these values into the Pythagorean theorem equation:
d^2 = (112 miles)^2 + (200 miles)^2
d^2 = 12544 miles^2 + 40000 miles^2
d^2 = 52544 miles^2
Taking the square root of both sides to solve for d:
d = √(52544 miles^2)
d ≈ 229.34 miles
Now, we can differentiate the equation with respect to time to find the rate at which d is changing:
2d * (dd/dt) = 0 + 0 (since the distances traveled by each car are constant)
Simplifying the equation:
2d * (dd/dt) = 0
dd/dt = 0 / (2d)
dd/dt = 0
Therefore, the rate at which the distance between the cars is increasing 4 hours later is 0 miles per hour. This means that after 4 hours, the distance between the cars will remain constant.