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

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  

    $147^2 = 0^2 + 2* a* 0.208$

=>  $a =51945 \ km/h^2$