An airplane accelerates with a constant rate of 3.0 m/s2 starting at a velocity of 21 m/s. If the distance traveled during this acceleration was 535 m, what is the final velocity?

The final velocity is 60.42 m/s

The parameters given in the question are;

initial velocity(u)= 21 m/s

acceleration(a)= 3.0 m/s²

distance(s)= 535 meters

final velocity(v)= ?

The final velocity can be calculated as follows

v²= u² +2as

v²= 21² + 2(3)(535)

v²= 441 + 3210

v²= 3651

v= √3651

v= 60.42

Hence the final velocity is 60.42 m/s

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60.42 m/s.

Explanation:

This problem can be solved using equation of motion

where v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance moved by the body to reach final velocity v from initial velocity u.

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given

u = 21m/s

a = 3.0 m/s2

s = 535 m

v, the final velocity we have to find

using the equation of motion

Thus, the final velocity of airplane is 60.42 m/s.

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