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 the airplane?

83.33 seconds.

Explanation:

We are given;

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

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

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

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

Step 2: Calculate the take off time

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.