A cessna aircraft has a lift-off speed of 120 km/h. what minimum constant acceleration does this require if the aircraft is to b
e airborne after a takeoff run of 280 m?
1 answer:
When you sketch the problem, it would look like that shown in the picture. The velocity the airplane should achieve must be 120 km/h in order for it to lift off the ground and take-off. Before this, it has to build up speed in order to reach the final velocity from rest. In rectilinear motion, one of the useful equations used is
2ax = vf² - vi²
For consistency, let's convert km/h to m/s.
120 km/h * (1000 m/1 km) * (1 h/3600 s) = 33.33 m/s
Substituting the values,
2a(280 m) = (33.33 m/s)² - 0²
a = 1.984 m/s
The minimum acceleration is 1.984 m/s.
2ax = vf² - vi²
For consistency, let's convert km/h to m/s.
120 km/h * (1000 m/1 km) * (1 h/3600 s) = 33.33 m/s
Substituting the values,
2a(280 m) = (33.33 m/s)² - 0²
a = 1.984 m/s
The minimum acceleration is 1.984 m/s.
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