An Airplane From Rest Accelerates On Runaway At 5.50 M/S Squared For 20.25 S Until It Finally Takes Of The Ground.What Is The Distance Covered Before

An airplane from rest accelerates on runaway at 5.50 m/s squared for 20.25 s until it finally takes of the ground.What is the distance covered before takeoff?

 

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Problem:

An airplane from rest accelerates on runaway at 5.50 m/s squared for 20.25 s until it finally takes of the ground.What is the distance covered before takeoff?

Given Data:

t (time) = 20.25 s  

Vi (initial velocity) = 0 m/s   (rest)

a (average acceleration) = 5.50 m/s²

d (distance covered) = ? unknown

Solution:

We can use the formula, d = 1/2at² since Vi is equal to 0 m/s

d = 1/2at²

   = 1/2 ( 5.50 m/s² ) × (20.25 s)²

  = 2.75 m/s² × 410.06 s²

  = 1127.67 m

Answer:

distance covered before takeoff is 1127.67 m