A car passes by a certain point at t = 0 hours heading east at 40 miles/hour. At t = 1 hour, another car passes by the same point heading north at 50 miles/hour. Which of the following functions equals the distance L (in miles) between the two cars as a function of t, for t is greater than or equal to 1?
a. L = 90t - 50
b. L = square root(4100t)
c. L = square root(4100t^2 + 5000t + 2500)
d. L = square root(4100t^2 - 5000t + 2500)
e. L = square root(90t-50)
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To find the distance L between the two cars as a function of t, we can use the concept of relative motion. Let's break down the problem step by step:
1. At t = 0 hours, the first car passes the point heading east at a constant speed of 40 miles/hour. Therefore, the position of the first car at any time t can be represented by the equation x1 = 40t, where x1 is the distance covered by the first car.
2. At t = 1 hour, the second car passes the same point heading north at a constant speed of 50 miles/hour. The position of the second car at any time t can be represented by the equation y2 = 50(t - 1), where y2 is the distance covered by the second car.
3. To find the distance L between the two cars, we can use the Pythagorean theorem. It states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
In this case, L is the hypotenuse, which is the distance between the two cars. The sides of the right triangle are the distances covered by each car. Therefore, we have:
L^2 = (x1)^2 + (y2)^2
L^2 = (40t)^2 + (50(t - 1))^2
4. Simplifying the equation, we get:
L^2 = 1600t^2 + 2500(t^2 - 2t + 1)
L^2 = 1600t^2 + 2500t^2 - 5000t + 2500
5. Combining like terms, we have:
L^2 = 4100t^2 - 5000t + 2500
6. Finally, taking the square root of both sides, we get:
L = square root(4100t^2 - 5000t + 2500)
Therefore, the correct answer is option d. L = square root(4100t^2 - 5000t + 2500).