Question:

An airplane is heading due south at a speed of 688 km/h. If a wind begins blowing from the southwest at a speed of 90.0 km/h (average), calculate

- the velocity (magnitude and direction) of the plane, relative to the ground, and
- how far from its intended position it will be after 11.0 min if the pilot takes no corrective action. [
*Hint:*First draw a diagram.]

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

Quick Answer:

- $628\textrm{ km/hr, }5.82^\circ \textrm{ East of South}$
- $16.5 \textrm{ km off course}$

### Transcript for this Giancoli solution

This is Giancoli Answers with Mr. Dychko. So this plane is directed downwards to the south with a velocity of 688 kilometers per hour and that's the plane with respect to the air's velocity. Now the air itself is moving with respect to the ground at an angle of 45 degrees from the southwest at 90 kilometers an hour and our question is what is their resulting velocity of the plane with respect to the ground and we find that by going velocity of plane with respect to the air plus velocity of air with respect to the ground and we see the inner subscripts are the same and this gives us the velocity with only the outer subscripts; the plane with respect to the ground in other words. We'll use the*cos*law to figure this out because that law is applicable when you have two sides of the triangle plus the enclosed angle— it works for any triangle, doesn't have to be a right triangle— and we know that this opposite side to the known angle is usually called

*c*squared equals

*a*squared plus

*b*squared minus 2 times the length of

*a*and the length of

*b*times the

*cos*of this angle,

*C*, 45 degrees in this case. So velocity of plane with respect to the ground is like little

*c*and it's gonna equal the square root then of all this stuff velocity of plane with respect to the air squared plus velocity of the air with respect to the ground squared minus 2 times those velocities multiplied together times

*cos*45. So that's the square root of 688 squared plus 90 squared minus 2 times 688 times 90 times

*cos*45 which is 627.60 kilometers per hour; that's the resultant velocity of the plane with respect to the ground. Then we have to find this angle

*Θ*and we'll use this

*sin*law to figure that out and the

*sin*law says that the

*sin*of some angle divided by the length of the side opposite the angle equals the

*sin*of any other angle divided by the length of the side opposite that angle. So we have

*sin Θ*over

*v AG*equals

*sin*45 over

*v PG*and we multiply both sides by

*v AG*here and that solves for

*sin Θ*. And then we'll take

*Θ*then is the inverse sign of this 90 kilometers an hour times

*sin*45 divided by

*v PG*which we have calculated—627.6 kilometers an hour— and that gives about 5.82 degrees and that's towards the east compared to south. So that's 628 kilometers an hour, 5.82 degrees east of south is the resultant velocity of the plane with respect to the ground. And the next question is, how far off course is the plane? So the plane's intended displacement is

*d A*, the actual displacement that it has is

*d B*and the difference between the two is this one; when you are subtracting vectors, you are going head to head so this is the first vector and this is the second vector and the difference is you put the tails together and connect the one head to the other head— that's subtraction of vectors— we could have also added the opposite and so on just to use the more familiar head to tail method maybe but this works fine. So this is intended displacement, this is actual displacement with respect to the ground and this is the difference which we have to calculate. Now the angle here is gonna be the same as the angle of the velocity because the displacement of the plane with respect to the ground is gonna be in the same direction as the velocity of the plane with respect to the ground. So the intended displacement

*d A*is straight down 688 times 11 minutes but we have to convert that into hours so divide by 60 or times by 1 hour for every 60 minutes and we end up with 688 times 11 over 60 gives 126.13 kilometers and after you take the air into account, the displacement with respect to the ground is gonna be 627.6 times 11 over 60 that gives 115 and then the difference we'll calculate using

*cos*law again. So this

*d AB*is the opposite this angle

*Θ*and we know both the sides that are beside this angle and so we can use the

*cos*law and it's gonna be the square root then of 90 kilometers... oops that's not 90 kilometers that's gonna be

*d A*which is 126.13 kilometers squared plus the 115.06 kilometers squared minus 2 times 126.13 times 115.06 times

*cos*of the angle which is 5.82 and then let's see what happens here. We have the square root of 126.13 plus 115.06 minus 2 times 126.13 times 115.06 times

*cos*of 5.82 and that gives 16.497; that's about 16.5 kilometers off course.

## Comments

Your answer for question a is listed above in units of meters/second, but the work shows that the units are actually kilometers/hour.

Hi clarkmg,

Thank you very much for noticing that. I've fixed the Quick Answer to reflect the video.

Best wishes with your studies,

Mr. Dychko

Dear Mr. Dychoko,

For part (a), I understand how you used the cosine rule to find the unknown vector C. However, I tried to solve the problem similar to what you did in problem 42 and divided the wind velocity into X and Y components. I used to pythagorean theorem to find the resultant vector but was not able to get the right answer. Can you please tell me why it is wrong to use this method.

Thank You

Hi chaegyunkang, thanks for the question. The method you describe works fine, so your sleuthing should look at how you implemented it. Pay careful attention to whether you subtracted the y-component of the velocity of the air with respect to the ground from the velocity of the plane with respect to the air (instead of adding it). Knowing that $\cos(45) = \dfrac{1}{\sqrt{2}}$, here's what the work will look like in your calculator:

$\Big(688 - \dfrac{90}{\sqrt{2}} \Big)^2 + \Big(\dfrac{90}{\sqrt{2}}\Big)^2 = \textrm{ some number}$

$\sqrt{ \textrm{ some number}} = 627.5953...$ which is the correct magnitude, and I'll leave the angle for you.

Hope this helps,

Mr. Dychko

My mistake was adding the vectors. Thank you so much.

You're welcome! That's a common mistake, so make sure you don't sweat it.