Question

an engineer is designing a runway. she knows that a plane, starting at rest, needs to reach a speed of 150mph 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 6000ft long runway?

Explanation:

Step1: Convert speed from mph to ft/s

We know that \(1\) mile \( = 5280\) ft and \(1\) hour \(=3600\) s.
The take - off speed \(v = 150\) mph.

$$v=\frac{150\times5280}{3600}\text{ ft/s}=220\text{ ft/s}$$

Step2: Use the formula for average speed in uniformly accelerated motion

Since the plane starts from rest (\(u = 0\)) and has a final speed \(v\), the average speed \(\bar{v}=\frac{u + v}{2}\).
Here, \(u = 0\) and \(v = 220\) ft/s, so \(\bar{v}=\frac{0 + 220}{2}=110\) ft/s.
The time \(t = 65.0\) s.

Step3: Calculate the distance using the formula \(d=\bar{v}\times t\)

$$d=110\times65.0 = 7150\text{ ft}$$

Step4: Compare the required distance with the given runway length

The required distance \(d = 7150\) ft and the runway length \(L = 6000\) ft.

Answer:

The plane needs to travel \(7150\) ft.
Since \(7150>6000\), the plane could not take - off on a \(6000\) ft long runway.