Question

An airplane has a takeoff speed of 62 m/s. assuming a constant acceleration of 1.7 m/s2, what is the minimum length of runway it needs for takeoff?

Answer 1

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:

$2aS=v^2 -u^2$

Re-arranging it and substituting the numbers, we find

$S=\frac{v^2-u^2}{2a}=\frac{(62 m/s)^2-0}{2(1.7 m/s^2)}=1131 m$