ground speed of 104=(26+ 78)knots north. If, on the other hand, the glider fies with a heading of 210 at 90 knots, the glider will stay at the edge of the cloud with a ground speed of 52 knots south. If the glider Hies into the wind at 240, with a speed equal to 52 knots, it will remain stationary with respect to the ground. Finally, the lowest speed at which the glider can fly to stay at the edge of the cloud is 45 knots and his will be possible for a heading of 270. The ground speed in this case will be 24 knots 210 240270° 330 45 90 Example Aircraft towing glider(Merriam) Airplane A is flying horizontally with a constant speed of 200km/h, and is towing a glider B. The glider is gaining altitude. The tow cable has a length r= 60m, and 0 is increasing at a constant rate of 5 degrees per econd. We want to determine the magnitude of the velocity, UG, and the acceleration, aG, of the glider for G The velocity of the glider will be UG=UA+UG/A. The velocity of the glider relative to A is best expressed using a polar coordinate system. Thus, we write UG/A=rer+reee Since the length of the cable is constant, vr=F=0, and 0= 5(2T/360)=0.0874 rad /sec, which gives vg=r0=5.236 m/s. We can now transform the velocity vector to cartesian coordinates to obtain v/A=5.236=5.236sin15°i+5.236cs15°j=1.3552i+5.0575jm/s v=((200/3.6)+1.3552)i+5.0575jm/s,vg=57.135m/s=205.7km/h For the acceleration aG =aG/A, since aA=0. We also know that 0=0, and, obviously, r=0, thus, -457enm/2,ac/s=4.57m/2ground speed of 104 = (26 + 78) knots north. If, on the other hand, the glider flies with a heading of 210o at 90 knots, the glider will stay at the edge of the cloud with a ground speed of 52 knots south. If the glider flies into the wind at 240o , with a speed equal to 52 knots, it will remain stationary with respect to the ground. Finally, the lowest speed at which the glider can fly to stay at the edge of the cloud is 45 knots and this will be possible for a heading of 270o . The ground speed in this case will be 24 knots. Example Aircraft towing glider (Merriam) Airplane A is flying horizontally with a constant speed of 200km/h, and is towing a glider B. The glider is gaining altitude. The tow cable has a length r = 60m, and θ is increasing at a constant rate of 5 degrees per second. We want to determine the magnitude of the velocity, vG, and the acceleration, aG, of the glider for the instant when θ = 15o . The velocity of the glider will be vG = vA + vG/A. The velocity of the glider relative to A is best expressed using a polar coordinate system. Thus, we write, vG/A = ˙rer + r ˙θeθ . Since the length of the cable is constant, vr = ˙r = 0, and ˙θ = 5(2π/360) = 0.0874 rad/sec, which gives vθ = r ˙θ = 5.236 m/s. We can now transform the velocity vector to cartesian coordinates to obtain, vG/A = 5.236eθ = 5.236 sin 15o i + 5.236 cos 15o j = 1.3552i + 5.0575j m/s , or, vG = ((200/3.6) + 1.3552)i + 5.0575j m/s, vG = 57.135 m/s = 205.7 km/h . For the acceleration aG = aG/A, since aA = 0. We also know that ¨θ = 0, and, obviously, ¨r = 0, thus, aG/A = −r ˙θ 2 er = −4.57er m/s2 , aG/S = 4.57 m/s2 . 4