- $35.8 \textrm{ km, } 45^\circ \textrm{ S of E}$
- $5.53 \textrm{ m/s}$
- $56.1 \textrm{ km, } 45^\circ \textrm{ S of E}$

### Solution video

# OpenStax College Physics, Chapter 3, Problem 52 (Problems & Exercises)

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*a*with respect to

*b*. And in this question, it’s the velocity of the airplane with respect to the wind that's important. So capital

*a*is little

*a*and capital

*b*is little

*w*. And we'll add these two velocities together, such that the subscripts are the one that we want listed first with respect to something called

*x*that we don't really care about. In this case that’ll be the ground, plus velocity of something else, which is

*x/i> with respect to the other thing that we want in our answer. So and this is going to be the velocity of the ground with respect to the wind. And when we do it this way, these inner subscripts so long as they're the same, they will cancel. And we'll be left with the velocity of the first subscript with respect to the second subscript. So, what we want then is the velocity of the air with respect to the wind, in which case we need to have the velocity of air with respect to the ground. Which we have directly given to us by the question. And then add to that the velocity of the ground with respect to the wind. And now that these inner subscripts, little*

*g*are the same, they cancel. Leaving us with the velocity of airplane with respect to the wind. And now, we’re not directly given the velocity of the ground with respect to the wind, we're given the velocity of the wind with respect to the ground. But the opposite of that, is going to be of the negative of it. So it’s gonna be the same vector, but just pointing in the opposite direction. This is the velocity of the ground with respect to the wind. And that is the negative of the velocity of the wind with respect to the ground. So that’s two meters per second co-linear with the velocity of the airplane with respect to the ground. So these two vectors can be placed along the same line, head to tail here. Here’s the velocity of the ground with respect to the wind. And we can see that this total then, which is from here to here. This resultant is the velocity of the airplane with respect to the wind and it’s gonna be just adding these two speeds together; So that’s 3.53 meters per second plus two meters per second. Which is 5.53 meters per second. And the displacement of the airplane with respect to the wind is going to be this speed of the airplane with respect to the wind multiplied by time, which is 56.1 kilometers and that is still 45 degrees south of the east.## Comments

Will you explain how you set up the relationships between air-ground, wind-ground etc? For example, how did you know 2.00m/s was wind-grd and not grd-wind. When I read any problem, how do I know which is which? and then how to do I set up the equation correctly? for example not: Vag=Vaw+Vwg?

Hello blue, thank you for the question. I totally understand how this is confusing. The subscripts are v_[thing moving][compared to what], which is to say, v_ag, for example, would be the velocity of the air (indicated by the *a*) compared to the ground (indicated by the letter *g*). When the problem says "He flew for 169 min at an average velocity of 3.53 m/s" the assumption is that the problem means 3.53 m/s compared to the ground. One can't avoid a speeding ticket by saying "I thought 60 mph was compared to the train moving along beside me, in which case I'm going only 10 mph officer!" The usual convention, in the absence of clarification, is that speeds are compared to the ground. The same goes for the headwind of 2.00 m/s: it's 2.00 m/s compared to the ground. I hope that helps.

All the best,

Shaun