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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.

Respuesta :

KimYurii

Answer:

[tex]\quad[/tex]0.0095 m/s²

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

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