Question

if you take off a velocity of an airplane on a runaway is 300 km/hr with an acceleration of 1 m/s^2 what is the take off time of the airplane

Answer 1

Given:

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

The acceleration of the plane, a=1 m/s²

To find:

The take-off time.

Step-by-step explanation:

The airplane starts from the rest and accelerates up to the take-off velocity. Thus the initial velocity of the airplane is u=0 m/s

The final velocity 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 take-off time of the plane.

On substituting the known values,