An airplane maintains a speed of 597 kmh relative to the air...

An airplane maintains a speed of 597 km/h relative to the air it is flying through as it makes a trip to a city 773 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 38.4 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 38.4 km/h to the east relative to the ground?

Transcript

00:02 So in this question we are given that there's an airplane that is flying north to a destination.

00:08 Let's call that destination is d and the original location of the plane has o.

00:15 And we are given that this north direction is the positive y direction and east direction is the positive x direction.

00:23 So basically the airplane must head north always in all cases.

00:28 And the velocity it maintains with respect to the air or the wind outside.

00:33 Is a vp of 597 kilometers per hour and the distance from its location to the destination is 773 kilometers in the north direction so we are given a few cases where there is wind in different directions and we have to find how much time it takes to reach the destination so in the first case we are given that there is a wind direction speed of 38 .4 kilometers per hour in the south direction.

01:05 So if you draw the free body diagram or the velocity diagram of the aircraft taken as a dot here, this is the vp and this is the v -w.

01:15 So we see that the resultant velocity will be vp minus vw because they are in opposite directions.

01:23 So the time taken here would be the distance which is 773 by the resultant velocity which is vp minus vw which is 597 minus vw which is 597 minus 38 .4 and this gives us that the time taken is 1 .38 hours in this case when there is a headwind down south of this velocity and the next part then asks us the same thing but it becomes a tailwind so this same velocity is now due north so time taken in this case would become because the resultant velocity in this case second case this becomes the vw and the result in velocity becomes addition of the two so time taken a little bit distance by the velocity which is 597 plus 38 .4 and this becomes 1 .22 hours which is of course shorter than the previous case and the wind was against the aircraft then the third part asks us to find out a case where the aircraft is supposed to head north but there is a wind direction due east and the velocity of the wind is the same 38 .4 or sorry the speed of the wind is the same the velocity is due east this time so at first glance it might look like it will do nothing to the aircraft because it is a 90 degrees with the motion of the aircraft but what happens is that if you if the aircraft does not change or does not adjust for this, it basically heads towards this direction and it does not head towards the north direction which is not correct.

03:20 So we want the resultant velocity to head north, which means that the vp will be at an angle of let's say, theta from this such that the vp x component will cancel out the wind component and the vp y component will be the one that will take it towards d...

Question

Please give Ace some feedback