Answer:
The take off time is equal to 41.66s.
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,
[tex]t=\dfrac{v}{a}\\\\t=\dfrac{83.33}{2}\\\\t=41.66\ s[/tex]
So, the take off time is equal to 41.66s.
