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3 years ago

14

A Cessna aircraft has a lift off speed of 147 km/h. What minimum constant acceleration does this require if the aircraft is to b

e airborne after a take off run of 208 m?

1 answer:

netineya [11]3 years ago

6 0

Answer:

The acceleration is $a =51945 \ km/h^2$

Explanation:

From the question we are told that

The lift up speed is $v = 147 \ km/h$

The distance covered for the take off run is $s = 208 m = 0.208 \ km$

Generally from kinematic equation we have that

$v^2 = u^2 + 2as$

Here u is the initial speed of the aircraft with value 0 m/ s give that the aircraft started from rest

So

$147^2 = 0^2 + 2* a* 0.208$

=> $a =51945 \ km/h^2$

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