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The taxiing time is 5S and the taxiing distance is 1000m. If the flight acceleration is approximately considered to remain unchanged, calculate the acceleration of the aircraft
"X = 0.5 at * t" so a = 80 meters per second
The taxiing time is 16.2s and the taxiing distance is 1200m,
If the acceleration of the aircraft is approximately constant, find: (1) the acceleration of the aircraft; (2) the speed of the aircraft at takeoff
Which formula transforms a = 2S / T
1. According to the meaning of the question, the plane does a uniform acceleration linear motion during takeoff
S = 1 / 2at ^ 2, that is, a = 2S / (T ^ 2) = 9.15m/s ^ 2
2. VT = V0 + at, that is, VT = 0 + at = 148m / s
The taxiing time is 16.2s and the taxiing distance is 1200m. If it is considered that the acceleration of the aircraft does not change approximately
The taxiing time is 16.2s and the taxiing distance is 1200m. If it is approximately considered that the acceleration of the aircraft does not change, calculate (1) the acceleration of the aircraft and (2) the speed of the aircraft when taking off
Hello
(1) According to the formula displacement = half * acceleration * time square, 1200 = half * acceleration a * (16.2) square
14 m / S ^ 2
(2) According to the formula VT = V0 + at, VT = 0 + 9.14 * 16.2 = 148.1m/s
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