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.
Explanation:
The initial velocity of the plane, u=0 m/s
The takeoff velocity of the plane 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 takeoff time of the plane.
On substituting the known values,
[tex]\begin{gathered} 83.33=0+2t \\ \Rightarrow t=\frac{83.33}{2} \\ =41.7\text{ s} \end{gathered}[/tex]
Final answer:
The takeoff time of the plane is 41.7 s
