Question

A Cessna 150 aircraft has a lift-off speed of approximately 125 km/h. What minimum constant acceleration does this require if the aircraft is to be airborne after a take-off run of 205 m? Answer in units of m/s 2 .

Answer 1

The minimum acceleration is $2.94 m/s^2$

Explanation:

The motion of the Cessna is a uniformly accelerated motion, so we can use the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity of the Cessna

u is the initial velocity

a is the acceleration

s is the distance covered when accelerated from u to v

Assuming the Cessna starts from rest, we have:

u = 0

The final velocity (lift-off speed) is:

$v=125 km/h \cdot \frac{1000 m/km}{3600s/h}=34.7 m/s$

While the maximum distance that the aircraft can cover is:

s = 205 m

Therefore, the minimum acceleration that the Cessna should have is:

$a=\frac{v^2-u^2}{2s}=\frac{(34.7)^2}{2(205)}=2.94 m/s^2$

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly