Two planes approach each other head-on. Each has a speed of 685 km/h, and they spot each other when they are i

Question

Two planes approach each other head-on. Each has a speed of 685 km/h, and they spot each other when they are initially 13.0 km apart. How much time do the pilots have to take evasive action?
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Answer 1

34.16 seconds (13km / 2 )/685 km/h =34.16 seconds

Answer 2

Since they are each moving at 685 km/h, the gap is closing at twice that rate (once for each plane).

Since you have 13 km, figure out how long it takes to get to 13 km at twice 685 km/hr.

As far as evasive action, the question presumes you can take evasive action up to the last millisecond, so I think it really means, how long until they collide?

Answer 3

Both planes are traveling at the same speed, so they will travel the same distance before colliding. 13 km/2= 6.5 km.
Each plane will travel 7.5 km before they crash.

Now just find the time it takes to travel 6.5 km at 685 km/h.
6.5 km/685 km/h=.009489 hours

.009489 hours=34.16 seconds
(.01094 hours * 60 min/hour*60 sec/min

The pilots have 34.16 seconds to take evasive action

Answer 4

I got 0.00948905 seconds as an answer.

How? Using a table, I came up with:

R(km/h)___T(h)___ D(km)
__685 _____ x ____ 685x - Plane one
__685 _____ x ____ 685x - Plane one

(sorry about the underlines. the answer scrunched everything together when I didn't put anything.)


685x+685x=13
1370x=13
x=0.00948905


This is because x, if you recall, represents time.

I hope that I have been able to help you =]