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.