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An airplane takes off at the airport with a velocity of 110 meters per second after traveling a distance of 125 meters. What was the acceleration?

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Final answer:

The acceleration of the airplane is 48.4 meters per second squared, calculated using the kinematic equation v^2 = u^2 + 2as with the given final velocity and distance.

Step-by-step explanation:

The subject of this question is Physics, and the grade level is High School. To calculate the acceleration of the airplane, we can use the kinematic equation:

v2 = u2 + 2as,

where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled. Since the airplane starts from rest, u = 0, v is given as 110 m/s (final velocity), and s is 125 meters. We can rearrange the equation to solve for acceleration (a):

a = (v2 - u2) / (2s)

Plugging in the values:

a = (1102 - 02) / (2 × 125)

= 12100 / 250

= 48.4 m/s2

Thus, the acceleration of the airplane is 48.4 meters per second squared.

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