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 271 m? Answer in units of m/s 2 .

Respuesta :

To calculate the minimum constant acceleration , we use the equation of motion with uniform acceleration,

[tex]v^{2} = u^{2} +2a S[/tex]

Here, v is final velocity, u is initial velocity, a is acceleration and s is distance.

Given  [tex]v = 125 \ km/h= \frac{125 \times 10^3 m}{60 \times 60 \ s}  =34 .72 \ m/s[/tex] and  [tex]s = 271 \ m[/tex].

Substituting these values with u = 0 (because initial velocity is zero) in above equation, we get

[tex](34.72 m/s) ^2 = 0 +  2 a \times 270 m \\\\ a = \frac{(34.72 m/s) ^2}{2 \times 270 m}  = 2.23 \ m/s^2[/tex]

Thus, the minimum constant acceleration is [tex]2.23 \ m/s^2[/tex]