An airplane maintains a speed of 573 kmh relative to the air...

An airplane maintains a speed of 573 km/h relative to the air it is flying through as it makes a trip to a city 703 km away to the north. (Assume north is the positive y-direction and east is the positive x-direction.) (a) What time interval is required for the trip if the plane flies through a headwind blowing at 35.7 km/h toward the south? h (b) What time interval is required if there is a tailwind with the same speed? h (c) What time interval is required if there is a crosswind blowing at 35.7 km/h to the east relative to the ground? h

Transcript

00:02 So in this question we are given that there is an aircraft airplane as shown here at point o let's say center of the origin of this whole diagram and it is headed towards a destination let's say d which is in the north direction as shown and we are told that there is a the distance between o and d 703 kilometers.

00:25 The velocity of the airplane is fixed with respect to the air in which it is flowing as 573 kilometers per hour.

00:33 And there is a wind velocity of 35 .7 which has a constant magnitude but it changes direction so we have to solve for various direction cases.

00:44 So the first part of this question is that we are asked to find out in various cases how much time will it take for the aircraft to reach from 0 to point d which is exactly north direction.

00:58 So the first case says that the vw is down south facing so southwards which means that if you draw free water diagram of our velocity vector diagram of the aircraft vp is the velocity with respect to the air but the air itself has a velocity of vw in the south direction so the relative velocity here will be vp minus vw in which case the time taken would be od divided by vp minus vw this will give us that it is 703 divided by 573 minus 35 .7, which is 1 .31 hours.

01:44 So this is the answer for your first part.

01:48 For the second part, we are given that there is a tailwind now which assists the aircraft because it's also heading north.

01:55 In this case, the vector will become aligned in the same direction.

02:03 So they will add up because the wind will assist the airplane speed, which is with respect to the wind itself.

02:09 So in this case the time taken becomes od by vp plus vw which becomes 703 by 573 plus 35 .7 which is 1 .15 hours.

02:30 Now in the third case we are told that there is a so this is the direction in which the aircraft should be heading and there is a crosswind of vw again but this is towards the east this time at 90 degrees with the north direction.

02:48 So if you think this looks simple because if vp is in this direction and vw is in this direction, there should be no problem.

02:56 Vp should just go on without any problem.

02:59 But the issue here is that if you have vp here and vw like this, you have a result in velocity which is here.

03:05 And then the aircraft is no more heading in the north direction.

03:08 It must go north towards d...

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