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A airplane accelerated from 59 m/s to 95 m/s in a distance of 123 meters what was its acceleration in m/s^2, assumed constant?

1 Answer

2 votes

Answer:

22.54 m/s^2

Step-by-step explanation:

vf = final velocity = 95 m/s

vi = initial velocity = 59 m/s

d = displacement = 123 m

a = acceleration, unknown

Use this kinematics equation to find a:

vf^2 = vi^2 + 2a*d

95^2 = 59^2 + 2*a*123

22.54 m/s^2 = a

Hope this helps!! :)

User DeDogs
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