Data:
u=0 m/s is the initial velocity of the plane
v=62 m/s is the final velocity of the plane (at which the plane takes off)
a=1.7 m/s^2 is the acceleration of the plane
To find the minimum distance S the plane needs to take off, we can use the following equation:
[tex]2aS=v^2 -u^2[/tex]
Re-arranging it and substituting the numbers, we find
[tex]S=\frac{v^2-u^2}{2a}=\frac{(62 m/s)^2-0}{2(1.7 m/s^2)}=1131 m[/tex]
