Given:
The takeoff velocity of the plane, v=300 km/hr
The acceleration of the plane, a=1 m/s²
To find:
The average velocity of the plane.
Explanation:
The initial velocity of the plane is u=0 m/s
The take-off 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]
The average velocity of an object having a constant acceleration is given by the sum of the initial and final velocity divided by two.
Thus the average velocity of the plane until it takes off is given by,
[tex]\begin{gathered} v_{av}=\frac{u+v}{2} \\ =\frac{0+83.33}{2} \\ =41.7\text{ m/s} \end{gathered}[/tex]
From the equation of motion,
[tex]v^2-u^2=2ad[/tex]
Where d is the takeoff distance.
On substituting the known values,
[tex]\begin{gathered} 83.33^2-0=2\times1\times d \\ \Rightarrow d=\frac{83.33^2}{2\times1} \\ =3471.9\text{ m} \end{gathered}[/tex]
Final answer:
Thus the average velocity of the plane is 41.7 m/s
Thus the takeoff distance is 3471.9 m
