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Problem 34 Hard Difficulty

The magnitude of a velocity vector is called speed. Suppose that a wind is blowing from the direction $\mathrm{N} 45^{\circ} \mathrm{W}$ at a speed of $50 \mathrm{~km} / \mathrm{h}$. (This means that the direction from which the wind blows is $45^{\circ}$ west of the northerly direction.) A pilot is steering a plane in the direction $\mathrm{N} 60^{\circ} \mathrm{E}$ at an airspeed (speed in still air) of $250 \mathrm{~km} / \mathrm{h}$. The true course, or track, of the plane is the direction of the resultant of the velocity vectors of the plane and the wind. The ground speed of the plane is the magnitude of the resultant. Find the true course and the ground speed of the plane.

Answer

$\bullet$ The ground speed: 267.34 $\mathrm{km} / \mathrm{hr}$
$\bullet$ The true course: E19.59'N.

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Video Transcript

for this problem, we have a plane that is traveling to the northeast with a given angle and but there is a wind that is blowing and it has um it comes from northwest With a velocity of or with the speed of 50 km/h. So we want to determine the true, the true course of the plane. And to that better we are going to call it t. And we also want to determine the ground, the speed of the plane and this is the speed is the magnitude both God and velocity. So to obtain team, we just simply need to soon the better of the wind plus the better of the plane. So to see this graphically, I'm going to draw the pictures, correspondent to the plane, the wind and also the resulting that George that we can obtain within the parallel Graham method self. Um if we draw the first one, it says that has a magnitude of 250 km/h. So are you gonna draw don't? And it also says that It has an angle of 60°. Nor eat. This means this angle in here, This is 50° and this is the actual p of the plane. No. We also know That the wind better has the magnitude of 50 km/h and is blowing from north due west. Um So it comes from here, putting in another colour, It comes from here and it has um angle um with respect to the northwest of 45 degrees. So we can extrapolate this back tour to his true position which is actually So 45 to the east. We can just translate it to here because they win. And the source of the wind is at northwest. But with its direction is to the soul uh is and to the sort of southeast is history direction. When we extrapolate that Petra. So we can see the first thing that we can do is to make sure that all of these angles um we're gonna call these um the live W. And all these angles are respect to the X angle to the ex exits. So that In this case we have that we have 60°. So we want this angle in here but we know that The white assets with access for 90°. So this in here will be 90 degrees minus 60 degrees. This is equal to 30 degrees. And with this one in here we see that This corresponds to 40 five degrees and all of this is 180° and swimming. All this also we will obtain plus 90 degrees. So we will have that the angle that the W. Or the better of the wind farms with the X. X. It is 45 plus 90 degrees Plus 180°. This will give us an angle of 350°. No, not that we know of the angles group respect to the X. Access. We want to obtain the resulted back tour from the son of the wind of the, from the wind back door plus them better off the plane. And I will give us the true course of the plane respect to the ground. So we can use the parallel ground mattered in the sense that we just struggle it. The people the p um that door and the don't live you better. And in this point we just obtain did this is the resulting back to work. T. Now the things that the thing that we are going to do is to determine the components of the wind vector and the playing better and then soon soup them, soup them with this equation and then obtain the true course of the plane. So doing that, we know from the figure that for this we have for the win back tour um our and the angle or we're gonna call it T to is off and for the other one in the plane We have that the magnitude is 250 km are our. And the angle we're going to call it all fell and it's 30° with respect to the ads to the X axis. So to obtain the company is very straightforward that we use the usual we we use the usual sine and Cosine functions to find the the company's. That's why we did we try to measure all the angles with respect to the X axis so that we can obtain the X. Component of the Dublin. You better as the magnitude Times the casino of Tita and in this case will have 50 kg/h Times Cosine of 350°. And for the case of the white company we will have deem the magnitude times the sine of Teton and this case is yeah so putting these values in the calculator and we're paying value of approximately for the first one 30 5.36 kilometers per hour and for the white part minus through it Carrying 5.36 km/h. Now we did the same with the plane back tour but in this case the angle is different the EDS company is is often so that is 250 km hour magazine of 30 so the they will these will give us 216 .51 km/h and the white part and this give us a value of 120 by kilometers are were now having all of this. Is is it too right those factors it will be the that's component last the why competent and these will give us and for the better of the plane we have so after doing this we need to some these two back doors to obtain the true of course better of the plane. So the true factor is equal to the picture of the wind times the back door sorry the better of the plane. So we know that we work on that and where when we are summoned to back towards with some company with the other company. You know with some adds company with ads company and white company with white company. So in this case it will be loss in the ads part. And oh in here I just forgot this. The white part of the wind vector is negative. So in here we will have uh minus Last 125 kilograms and kilometers per hour. And now assuming all of this, we obtained a true course of the plane is 200 51.87 km/h in the x component glass 80 89 0.54 Sheila matures per hour in the white part. So this is the true course. And now we are as also um the ground is speed. So since this is the velocity we see from the ground, the speed is just the magnitude of that better or so to find the magnitude which take the square root of the square of each company. This number of each company to the square. So we will have yes. So this give us a value of approximately 206, the seven point 35 km/h. And this is the speed of the plane measure from the ground