Question

A Cessna aircraft has a lift off speed of 147 km/h. What minimum constant acceleration does this require if the aircraft is to be airborne after a take off run of 208 m?

Answer 1

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

Step-by-step 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`