Answer:
The take off time is equal to 41.66s.
Step-by-step explanation:
Given that,
The initial speed of the airplane, u = 0
Acceleration, a = 2 m/s²
Let v is the take off time. Using first equation of motion,
v = u +at
Put v = 300 km/hr = 83.3 m/s
So,
![t=(v)/(a)\\\\t=(83.33)/(2)\\\\t=41.66\ s](https://img.qammunity.org/2022/formulas/physics/college/uaqq9j6i03xviwizbrzsrvmh2qiw5gxr4t.png)
So, the take off time is equal to 41.66s.