Two surveyors with two-way radios leave the same point at 9:00am., one walking due south at 2 mi/hr and the other due west at 5 mi/hr. How long can they communicate with one another if each radio has a maximum range of 10 miles?
Set up a right triangle. After 1 hour, the triangle will have one leg of 2, the other 5, and a hypotenuse of sqrt(4+25) = 5.385
Now, after t hours, the sides will be 2t and 5t, and the hypotenuse will be 5.385t
We want to find out when they are 10 mi apart; that's just the length of the hypotenuse.
5.385t = 10
t = 1.857 hrs = 1 hr 52 min
To find out how long the surveyors can communicate with each other, we need to determine when the distance between them becomes greater than the maximum range of the radios.
Let's assume that t hours have passed since 9:00 am. The surveyor walking due south will have traveled 2t miles, while the surveyor walking due west will have traveled 5t miles.
Using the Pythagorean theorem, we can find the distance between them at time t:
Distance^2 = (Distance South)^2 + (Distance West)^2
Now we can plug in the distances:
Distance^2 = (2t)^2 + (5t)^2
Simplifying:
Distance^2 = 4t^2 + 25t^2
Distance^2 = 29t^2
Taking the square root of both sides:
Distance = sqrt(29t^2)
Distance = sqrt(29) * t
We want to find the time when the distance is equal to the maximum range of the radios, which is 10 miles:
10 = sqrt(29) * t
Now we can isolate t by squaring both sides:
100 = 29t^2
Divide both sides by 29:
t^2 = 100 / 29
Taking the square root of both sides:
t = sqrt(100 / 29)
Using a calculator, we can find the approximate value of t:
t ≈ 1.96 hours
Therefore, the surveyors can communicate with each other for approximately 1.96 hours or 1 hour and 57 minutes.