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.
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
[tex]\begin{gathered} v=300\times\frac{1000}{3600} \\ =83.33\text{ m/s} \end{gathered}[/tex]
From the equation of motion,
[tex]v=u+at[/tex]
Where t is the take-off time of the plane.
On substituting the known values,
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