Two Planes approach each other head-on. One plane has a speed of 800 km/h and the other has a speed of 900 km/h and they spot each other when they are initially 15 km apart. How much time in seconds do the pilots have to take evasive action?
they are headed toward each other at
(800+900) /(3600) km/s
= .4722 km/second
15 km = .4722 t
t = 31.8 seconds
To find out how much time the pilots have to take evasive action, we can calculate the time it takes for the two planes to meet.
First, let's convert the speeds of the planes from km/h to m/s for consistency. We know that 1 km/h is equivalent to 1000 m/3600 s, so the speed of the first plane is 800 km/h * 1000 m/3600 s = 222.22 m/s, and the speed of the second plane is 900 km/h * 1000 m/3600 s = 250 m/s.
Let's denote the time it takes for the planes to meet as 't'. During this time, the first plane will have traveled a distance of 222.22 m/s * t, and the second plane will have traveled a distance of 250 m/s * t.
Since the planes are initially 15 km apart, which is equivalent to 15,000 meters, the sum of the distances traveled by the two planes must equal the initial distance between them:
222.22 m/s * t + 250 m/s * t = 15,000 m
Combining like terms, we get:
472.22 m/s * t = 15,000 m
Now we can solve for 't':
t = 15,000 m / 472.22 m/s ≈ 31.77 seconds
Therefore, the pilots have approximately 31.77 seconds to take evasive action.