Question

if it take off a lost city of an airplane on a run away is 300 km/hr with an acceleration of 2 m/s^2 what is a takeoff time of the airplane

Answer 1

Given:

The take-off velocity of the plane, v=300 km/hr

The acceleration of the plane, a=2 m/s²

To find:

The takeoff time of the airplane.

Step-by-step explanation:

The initial velocity of the plane, u=0 m/s

The takeoff velocity of the plane in m/s is,

`\begin{gathered} v=300*(1000)/(3600) \\ =83.33\text{ m/s} \end{gathered}`

From the equation of motion,

`v=u+at`

Where t is the takeoff time of the plane.

On substituting the known values,

`\begin{gathered} 83.33=0+2t \\ \Rightarrow t=(83.33)/(2) \\ =41.7\text{ s} \end{gathered}`

Final answer:

The takeoff time of the plane is 41.7 s