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A small turbo-prop commuter airplane, starting from rest on a Lansing airport runway, accelerates for 22.5s before taking off. Its speed at takeoff is 53.0 m/s (119 mi/hr). Calculate the acceleration of the plane, in g's, assuming it remains constant. (i.e., divide the acceleration in m/s2 by 9.81 m/s2).In the problem above, how far did the plane move while accelerating for 22.5 s?

Respuesta :

Answer:

s = 596.25 m

Explanation:

given,

initial speed, u = 0 m/s

final speed, v = 53 m/s

time, t = 22.5 s  

acceleration of the plane = ?

[tex]a = \dfrac{v-u}{t}[/tex]

[tex]a = \dfrac{53-0}{22.5}[/tex]

     a = 2.36 m/s²

using equation of motion

[tex]s = u t + \dfrac{1}{2}at^2[/tex]

[tex]s =\dfrac{1}{2}\times 2.36\times 22.5^2[/tex]

s = 596.25 m

Hence, the distance traveled by the plane is equal to 596.25 m

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