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ludmilkaskok [199]
2 years ago
7

If the take off velocity of an airplane on a runway is 300 km /hr with an acceleration of 1 m/s2. What is the take off time of t

he airplane?
Chemistry
1 answer:
Rama09 [41]2 years ago
5 0
<h3>Answer:</h3>

83.33 seconds.

<h3>Explanation:</h3>

<u>We are given;</u>

  • Take off velocity as 300 km/hr
  • Acceleration as 1 m/s²

We are required to calculate the take off time of the airplane.

<h3>Step 1: Convert velocity from km/hr to m/s </h3>

We are going to use the conversion factor.

The conversion factor is 3.6 km/hr per m/s

Therefore;

Velocity = 300 km/hr ÷ 3.6 km/hr per m/s

             = 83.33 m/s

<h3>Step 2: Calculate the take off time</h3>

We know that;

v = u + at

where, u is the initial velocity, v the final velocity, a the acceleration and t is time.

But, initial velocity is Zero

Therefore;

83.33 m/s = 1 m/s² × t

Thus;

time = 83.33 m/s ÷ 1 m/s²

       = 83.33 seconds

Therefore, the take off time is 83.33 seconds.

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