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.

We are given;

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

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

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.

the answer is b & c i believe : )