An airplane flies eastward and always accelerates at a constant rate. At one position along its path, it has a velocity of 30.7 m/s. It then flies a further distance of 41300 m, and afterwards, its velocity is 49.1 m/s. Find the airplane's acceleration.

0.0178 m/s²

Explanation:

From equation of motion,

v² = u² = 2as ... Equation 1

Where v = final velocity of the airplane, u = initial velocity of the airplane, a = acceleration of the air plane, s = distance covered by the airplane.

make a the subject of the equation,

a = (v² - u²) / 2s ... Equation 2

Given: v = 49.1 m/s, u = 30.7 m/s, s = 41300 m

Substitute into equation 2

a = (49.1² - 30.7²) / (2*41300)

a = (2410.81-947.49) / 82600

a = 1463.32/82600

a = 0.0178 m/s²

Hence the airplane's acceleration = 0.0178 m/s²