Question

A certain aircraft has a liftoff speed of 112 km/h.(a) What minimum constant acceleration does the aircraft require if it is to be airborne after a takeoff run of 280 m? m/s2(b) How long does it take the aircraft to become airborne? s

Answer 1

Answer:

(a) 1.73 m/s²

(b) 18 s

Explanation:

Part a)

To calculate the minimum constant acceleration, we will use the following equation

$v_f^2=v_0^2+2ax$

Where vf is the final velocity, v0 is the initial velocity, a is the acceleration and x is the distance traveled.

Now, convert 112 km/h to m/s as follows

$112\text{ km/h}\times\frac{1000\text{ m}}{1\text{ km}}\times\frac{1h}{3600\text{ s}}=31.11\text{ m/s}$

Then, we can replace vf = 31.11 m/s, v0 = 0 m/s, x = 280 m and solve for a, so

$\begin{gathered} 31.11^2=0^2+2a(280) \\ 967.9=560a \\ \\ \frac{967.9}{560}=a \\ \\ 1.73\text{ m/s}^2=a \end{gathered}$

Therefore, the acceleration is 1.73 m/s².

Part b)

To calculate the time, we will use the following equation

$x=v_0t+\frac{1}{2}at^2$

Replacing x = 280 m, v0 = 0 m/s, and a = 1.73 m/s² and solving for t, we get:

$\begin{gathered} 280=0t+\frac{1}{2}(1.73)t^2 \\ \\ 280=0.86t^2 \\ \\ \frac{280}{0.86}=t^2 \\ \\ 324=t^2 \\ \\ \sqrt{324}=t \\ 18s=t \end{gathered}$

Then, the aircraft becomes air airborne after 18 seconds