The plane will need 330.63m of runway for takeoff.
Why?
Since we already know the desired speed (46 m/s) of the plane and we know its acceleration (constant acceleration, 3.2 m/s2), we can calculate how much runway is needed for takeoff.
We can solve the problem using the following formula:
[tex]V^{2}=V_{o}^{2} +2a*d[/tex]
[tex]d = \frac{(V^{2} - Vo^{2} )}{2a}[/tex]
Where,
V is the final speed
Vo is the initial speed (which is equal to 0 assuming that the plane was in rest)
a is the acceleration of the movement.
d is the distance.
Now, substituting the given information and calculating, we have:
[tex](46\frac{m}{s} )^{2}=(0)^{2} +2*(3.2\frac{m}{s^{2} }) *d\\\\2116\frac{m^{2}}{s^{2}}=2*(3.2\frac{m}{s^{2} })*d\\\\2116\frac{m^{2}}{s^{2}}=6.4\frac{m}{s^{2} }*d\\\\d=\frac{2116\frac{m^{2}}{s^{2}}}{6.4\frac{m}{s^{2} }}=330.63m[/tex]
Hence, we have that the plane will need 330.63m of runway for takeoff.
Have a nice day!
