Question

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?

Answer 1

Answer:

60.42 m/s.

Explanation:

This problem can be solved using equation of motion

$V^2 = u^2 + 2as$

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

$V^2 = u^2 + 2as\\V^2 = 21^2 + 2*3*535\\V^2 = 441 + 3210 = 3651\\\sqrt{v^2} = \sqrt{3651} \\v = 60.42$

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