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

1 answer

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.