Question

An engineer is designing a runway. She knows that a plane, starting at rest, needs to reach a speed of 180mph at take-off. If the plane can reach this take-off speed in 65.0s, how far will the plane travel before take-off. Assume the plane increases speed steadily.
How far does the plane need to travel during take-off (in given units)?
ft
Could the plane take off on a 10000ft long runway?

Answer 1

Answer:

The plane would need to travel at least $8,\!580\; {\rm ft}$ ($8.58 \times 10^{3}\; {\rm ft}$.)

The $10,\!000\; {\rm ft}$ runway should be sufficient.

Explanation:

Convert unit of the the take-off velocity of this plane to $\rm ft\cdot s^{-1}$:

$\begin{aligned}v &= 180\; {\rm mph} \\ &= 180\; {\rm mi \cdot hrs^{-1}} \times \frac{1\; {\rm hrs}}{3600\; {\rm s}} \times \frac{5280\; {\rm ft}}{1\; {\rm mi}} \\ &= 264\; {\rm ft \cdot s^{-1}}\end{aligned}$.

Initial velocity of the plane: $u = 0\; {\rm ft \cdot s^{-1}}$.

Take-off velocity of the plane $v =264\; {\rm ft\cdot s^{-1}}$.

Let $x$ denote the distance that the plane travelled along the runway. Since acceleration is constant but unknown, make use of the SUVAT equation $x = ((u + v) / 2) \, t$.

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 $t$ times average velocity $(u + v) / 2$.

The distance that the plane need to cover would be:

$\begin{aligned}x &= \left(\frac{u + v}{2}\right)\, t \\ &= \frac{0\; {\rm ft \cdot s^{-1}} + 264\; {\rm ft \cdot s^{-1}}}{2} \times 65.0\; {\rm s} \\ &= 8.58\times 10^{3}\; {\rm ft}\end{aligned}$.