Question

An airplane accelerates from a velocity of 25.88 m/s at a constant rate of 4.21 m/s^2 over 590.39 m. What is its final velocity?

Answer 1

The final velocity is 75.1 m/s

What is Acceleration ?

Acceleration is the velocity change per time taken. It is measured in meter per square second.

Given that an airplane accelerates from a velocity of 25.88 m/s at a constant rate of 4.21 m/s^2 over 590.39 m. What is its final velocity?

The given parameters are;

• Acceleration a = 4.21 m/s^2

• Distance s = 590.39 m

• Initial velocity u = 25.88 m/s

• Final velocity v = ?

Let us use 3rd equation of motion. That is; v² = u² + 2as

Substitute all the parameters into the above equation

v² = 25.88² + 2 × 4.21 × 590.39

v² = 669.8 + 4971.08

v² = 5640.88

v = √5640.88

v = 75.11 m/s

Therefore, its final velocity is 75.1 m/s approximately

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