AVIA 305 Test Questions with Verified Answers Latest Update 2025 Already Passed

AVIA 305 Test Questions with Verified Answers Latest Update 2025 Already Passed An airplane is towing a glider to altitude. The tow rope is 20° below the horizontal and has a tension force of 300 lb exerted on it by the airplane. Find the horizontal drag of the glider and the amount of lift that the rope is providing to the glider. Sin 20° = 0.342; cos 20 °= 0.940. - Answers Drag: Cos 20° = D/T (Rearrange) D = Cos 20° X Tension D = 0.940 X 300lbs D = 282lbs Lift: Sin 20° = L/T (Rearrange) Lift = Sin 20° X Tension Lift = 0.342 X 300 lbs Lift = 102.6 lbs A jet airplane is climbing at a constant airspeed in no‐wind conditions. The plane is directly over a point on the ground that is 4 statute miles from the takeoff point and the altimeter reads 15,840 ft. Find the tangent of the plane's climb angle and the distance that it has flown through the air. - Answers Find the tangent of the plane's climb angle: Convert Altitude Into Miles: 15,840 ft / 5280ft = 3sm Tan = Height / Distance Tan = 3sm / 4sm Tan = 0.75 Find the Distance that it has flown through the air: Da = √ Distance² + Height² Da = √ 4sm² + 3sm² Da = √ 4 x 4 + 3 x 3 Da = √ 25sm Da = 5sm Find the distance (S) and the force (F) on the seesaw fulcrum shown in the figure. Assume that the system is in equilibrium. - Answers For the system to be in equilibrium, the moments on the left and right of the fulcrum must be equal. Find the Distance: Moment = Force (W) X Distance Weight 1 X S = Weight 2 X (Distance - S) (20lbs)S = 10lbs (24ft - S) 20S = (10)(24) + (10)(-s) + 10s [20S= 240 - 10s] +10s 30S = 240 /30 [30s = 240] /30 S = 8 Find the Fulcrum: Since the system is in equilibrium, the force (F) is equal to the sum of the two weights F = 20 + 10 F = 30 The airplane in Problem 2 starts from a brakes‐locked position on the runway. The airplane takes off at an airspeed of 200 fps. Find the time for the aircraft to reach takeoff speed. - Answers (Use answer from problem 2 as acceleration rate) A = 12 fps² Vo = 0 V1 = 200fps V = Vo + at (Rearrange) T = V - Vo / A T = 200fps - 0 fps / 12fps² T = 200fps / 12fps² T = 16.67 Seconds Under no‐wind conditions, what takeoff roll is required for the aircraft in Problem 6? - Answers S = Speed² - Initial speed² / 2 x acceleration rate S = (200fps)² - (0fps)² / 2 X 12fps² 200 x 200 0 2x24 S = 40,000fps² / 24fps² S = 1,667.7 ft Upon reaching a velocity of 100 fps, the pilot of the airplane in Problem 6 decides to abort the takeoff and applies brakes and stops the airplane in 1000 ft. Find the airplane's deceleration. - Answers S = V² - Vo² / 2a (Rearrange) A = V² - Vo² / 2 x s Vo Initial speed = 100fps V2 Desired Speed = 0 fps Distance = 1,000ft