# An engineer is designing the runway for an airport. Of the planes that will use the airport, the lowest acceleration rate is likely to be 3 m/s2. The takeoff speed for this plane will be 65 m/s. Assuming this minimum acceleration, what is the minimum allo

To find the minimum length of the runway required for the plane to reach its takeoff speed, we can use the kinematic equation that relates acceleration, initial velocity, final velocity, and distance covered:
\[
v^2 = u^2 + 2as
\]
where:
- \( v \) is the final velocity (takeoff speed, in this case),
- \( u \) is the initial velocity (assumed to be 0 for takeoff),
- \( a \) is the acceleration,
- \( s \) is the distance (length of the runway).
Given:
- Final velocity \( v = 65 \, \text{m/s} \)
- Initial velocity \( u = 0 \, \text{m/s} \)
- Acceleration \( a = 3 \, \text{m/s}^2 \)
We can rearrange the equation to solve for \( s \):
\[
s = \frac{v^2 - u^2}{2a}
\]
Substituting the known values:
\[
s = \frac{(65 \, \text{m/s})^2 - (0 \, \text{m/s})^2}{2 \times 3 \, \text{m/s}^2}
\]
Calculating \( (65 \, \text{m/s})^2 \):
\[
(65 \, \text{m/s})^2 = 4225 \, \text{m}^2/\text{s}^2
\]
Now substituting this back into the equation for \( s \):
\[
s = \frac{4225 \, \text{m}^2/\text{s}^2}{2 \times 3 \, \text{m/s}^2} = \frac{4225 \, \text{m}^2/\text{s}^2}{6 \, \text{m/s}^2}
\]
Calculating \( s \):
\[
s = \frac{4225}{6} \approx 704.17 \, \text{m}
\]
Thus, the minimum allowed length for the runway is approximately **704.17 meters**.