Question

6) Taking off is only one part of an airplane’s journey. The next part is landing.
Landing presents the same problem that taking off does from an aircraft carrier, limited space.
A fighter jet can fly at a cruising speed of around 270 m/s. If the average braking acceleration of that jet is 10.47 m/s^2, how much distance would the jet need in order to come this cruising speed to a complete stop. Show all work

Answer 1

To determine the distance required for the fighter jet to come to a complete stop, we can use the kinematic equation that relates initial velocity, acceleration, final velocity, and distance:

\[
v^2 = u^2 + 2a s
\]

Where:
- \(v\) = final velocity (0 m/s, since the jet comes to a stop)
- \(u\) = initial velocity (270 m/s)
- \(a\) = acceleration (which will be negative since it's a deceleration, so -10.47 m/s²)
- \(s\) = distance (the value we want to find)

We can rearrange the equation to solve for \(s\):

\[
s = \frac{v^2 - u^2}{2a}
\]

Substituting the known values into the equation:

\[
s = \frac{0^2 - (270)^2}{2 \times (-10.47)}
\]

Calculating \(u^2\):

\[
(270)^2 = 72900
\]

Now plugging that value into the equation:

\[
s = \frac{0 - 72900}{2 \times (-10.47)}
\]
\[
s = \frac{-72900}{-20.94}
\]
\[
s = \frac{72900}{20.94}
\]
\[
s \approx 3475.54 \, \text{m}
\]

Thus, the distance required for the fighter jet to come to a complete stop is approximately \(3475.54\) meters.