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Question: 

Two planes approach each other head-on. Each has a speed of 780 km/h, and they spot each other when they are initially 10.0 km apart. How much time do the pilots have to take evasive action?

Source: Giancoli, Douglas C., Physics: Principles with Applications, 7th Edition, 2014.

Quick Answer: 

$23.1 \textrm{ s}$

Giancoli 7th Edition, Chapter 3, Problem 41

(2:43)

Chapter 3, Problem 41 is solved.

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Transcript for this Giancoli solution

This is Giancoli Answers with Mr. Dychko. These planes are approaching each other at a speed of 780 kilometers an hour and they are initially 10 kilometers apart and to find the amount of time that they have to take evasive action, we want to know what the relative speeds of the planes are. So the pilot in this plane perceives this plane approaching at what speed? Well, we need to find the velocity of plane 1 with respect to plane 2. So we need to add the velocity of the plane 1 with respect to the ground which is what we are given— 780 kilometers an hour— plus the velocity of the ground with respect to plane 2 and that's kind of an odd way of saying this velocity is a velocity of the plane with respect to the ground and then that's typically what we quote. But we want to write it this way so that the inner subscripts are the same and they all cancel, in other words, leaving us with the velocity of the plane 1 with respect to plane 2. So to switch this velocity of plane 2 with respect to ground into ground with respect to P 2, we'll use this relation that you can take the negative and then flip the subscripts. So velocity of the ground with respect to plane 2 is the negative of the velocity of plane 2 with respect to ground. Now since the right is a positive direction velocity of plane 2 with respect to ground is actually negative 780— I didn't write a negative sign there because I already got an arrow point to the left to indicate direction but v P 2G is negative 780 kilometers an hour and then we take the negative of that to get the velocity of the ground with respect to plane 2. So the ground appears to be moving to the right at 780 kilometers an hour if you were to be in the reference frame of plane 2. Okay. So that's why you have positive 780 there. And that means the velocity of plane 1 with respect to plane 2 is 780 plus 780 which is 1560 kilometers an hour so they are approaching each other at this rate. And how long will it take with this relative speed to close the gap of 10 kilometers? And the time will be the distance divided by the speed so it's 10 kilometers divided by 1560 kilometers an hour which is this many hours and multiply that 0.0064103 hours by 3600 seconds per hour and the hours cancel leaving us with 23.1 seconds the amount of time for the pilot's to take evasive action.