Answer:
The plane would need to travel at least [tex]8,\!580\; {\rm ft}[/tex] ([tex]8.58 \times 10^{3}\; {\rm ft}[/tex].)
The [tex]10,\!000\; {\rm ft}[/tex] runway should be sufficient.
Explanation:
Convert unit of the the take-off velocity of this plane to [tex]\rm ft\cdot s^{-1}[/tex]:
[tex]\begin{aligned}v &= 180\; {\rm mph} \\ &= 180\; {\rm mi \cdot hrs^{-1}} \times \frac{1\; {\rm hrs}}{3600\; {\rm s}} \times \frac{5280\; {\rm ft}}{1\; {\rm mi}} \\ &= 264\; {\rm ft \cdot s^{-1}}\end{aligned}[/tex].
Initial velocity of the plane: [tex]u = 0\; {\rm ft \cdot s^{-1}}[/tex].
Take-off velocity of the plane [tex]v =264\; {\rm ft\cdot s^{-1}}[/tex].
Let [tex]x[/tex] denote the distance that the plane travelled along the runway. Since acceleration is constant but unknown, make use of the SUVAT equation [tex]x = ((u + v) / 2) \, t[/tex].
Notice that this equation does not require the value of acceleration. Rather, this equation make use of the fact that the distance travelled (under constant acceleration) is equal to duration [tex]t[/tex] times average velocity [tex](u + v) / 2[/tex].
The distance that the plane need to cover would be:
[tex]\begin{aligned}x &= \left(\frac{u + v}{2}\right)\, t \\ &= \frac{0\; {\rm ft \cdot s^{-1}} + 264\; {\rm ft \cdot s^{-1}}}{2} \times 65.0\; {\rm s} \\ &= 8.58\times 10^{3}\; {\rm ft}\end{aligned}[/tex].
