Question

An airplane flies eastward and always accelerates at a constant rate. At one position along its path, it has a velocity of 32.7 m/s . It then flies a further distance of 48500 m , and afterwards, its velocity is 44.7 m/s . Find the airplane's acceleration.

Answer 1

`\quad`0.0095 m/s²

Step-by-step explanation:

We have,

• Initial velocity (u) = 32.7 m/s
• Final velocity (v) = 44.7 m/s
• Distance travelled (s) = 48500 m

We are asked to calculate the airplane's acceleration.

By using the third equation of motion,

➝ v² - u² = 2as

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• s denotes distance

➝ (44.7)² - (32.7)² = 2 × a × 48500

➝ 1998.09 - 1069.29 = 97000 × a

➝ 928.8 = 97000a

➝ 928.8 ÷ 97000 = a

➝ 0.0095 m/s² = a