1. **State the problem:** We are asked to find the minimum length of a runway needed for a plane to take off. We know the plane's takeoff speed is $65 \text{ m/s}$, and the minimum acceleration during takeoff is $3 \text{ m/s}^2$. 2. **Recall the kinematic equation:** To find the minimum runway length $d$, we use the formula relating initial velocity, final velocity, acceleration, and distance: $$v^2 = u^2 + 2ad$$ Here, $u$ is the initial velocity (which is 0 since the plane starts from rest), $v$ is the final velocity ($65 \text{ m/s}$), $a$ is the acceleration ($3 \text{ m/s}^2$), and $d$ is the distance (runway length). 3. **Plug in known values:** Since the plane starts from rest, $u = 0$, so $$65^2 = 0 + 2 \times 3 \times d$$ $$4225 = 6d$$ 4. **Solve for $d$:** $$d = \frac{4225}{6} = 704.166... \text{ m}$$ 5. **Conclusion:** The minimum allowed length for the runway is approximately $704.17$ meters.