If a fighter jet touches the runway at 220km/h, then decides to brake, and finally comes to a stop at 31 seconds later.

What is the shortest runway the fighter jet can land, assuming constant deceleration starting at touchdown?

Is the answer 570 meters or am I just totally off

1) change speed to m/s

61.1m/s check that.

avgspeed=(vf+vi)/2=30.5m/s

distance=avgspeed*time

I get considerably more than you

Remember, under constant deaccleration to a stop, the average speed is 1/2 the initial speed.

To find the shortest runway length, we can use the physics formula:

\[v_f^2 = v_i^2 + 2a d\]

where:
- \(v_f\) is the final velocity (0 km/h since the jet comes to a stop),
- \(v_i\) is the initial velocity (220 km/h),
- \(a\) is the deceleration (we need to find this),
- \(d\) is the distance (runway length) covered by the jet.

In this case, \(v_f = 0\) km/h and \(v_i = 220\) km/h.

To solve for \(a\) and \(d\), we need another piece of information - the time taken for the jet to stop, \(t\), which is given as 31 seconds.

The formula connecting acceleration, velocity, and time is:

\[v_f = v_i + a t\]

From this equation, we can isolate \(a\) and plug it into the first formula to find \(d\).

Isolating \(a\):

\[a = \frac{{v_f - v_i}}{{t}}\]

Substituting the known values:

\[a = \frac{{0 - 220}}{{31}}\]

\[a = -\frac{{220}}{{31}}\]

Now we can use the first formula with \(v_f = 0\), \(v_i = 220\), and \(a = -\frac{{220}}{{31}}\) to find \(d\).

\[0 = (220)^2 + 2\left(-\frac{{220}}{{31}}\right) d\]

\[0 = 48400 - \frac{{440}}{{31}} d\]

Solving for \(d\):

\[\frac{{440}}{{31}} d = 48400\]

\[d = \frac{{48400 \times 31}}{{440}}\]

\[d \approx 3446.36\]

So, the shortest runway length the fighter jet can land on, assuming constant deceleration starting at touchdown, is approximately 3446.36 meters.

Therefore, your answer of 570 meters is incorrect.