HP question

[Pilotpaul787]

Hi all, I am ready to take my HP and my comms exam both on Thursday this week. I just have one question however, I've come across a question a few times during practice exams its the only thing bugging me about the HP exam, I'll give you a copy of the question below and if someone could let me know how they would work it out Id appreciate it. I can't find an adequate example in my text book. The question is;

With an in-flight visibility of 4 km a light aircraft flying at 120kts sees a fast jet at maximum visual range. The light aircraft pilot assumes that the fast jet is flying at 450 knots what time would be left to the pilot of either aircraft to take avoiding action?
a) about 7 seconds
b) about 10 seconds
c) about 14 seconds
d) about 17 seconds

Correct answer is c) about 14 seconds

With a closing speed of 570 knots and a range of 4km I can't seem to work out the calculation needed for such a question. Thanks in advance.

Paul

p.s anyone who answered my question on met a few weeks back thanks for your help I passed the exam about 2 weeks ago:ok:

[BackPacker]

Simple math. But don't confuse the km and knots!

The closure speed is 120 plus 450 knots = 570 knots = 1026 km/hr = 17.1 km/minute = 0.285 km/sec.

Inflight viz is 4 km. At 0.285 km/s, we have a reaction time of 4 / 0.285 = 14 seconds.

That's of course assuming the avoiding action itself will take zero seconds and that you were looking at the target as and when it came into view, and recognised it for what it was.

Bit of an academic question really.

[Pilotpaul787]

Thanks Backpacker I see now

Paul

[stickandrudderman]

Bit of an academic question really.

As are 80% of the PPL questions!