The problem asks to determine the minimum required runway length for a plane based on the following information:

• The position of the plane as a function of time $x(t)$ is given by: $x(t) = 1.3t^2 \text{ meters}$
• The minimum takeoff speed is $v = 33.8 \, \text{m/s}$.

To solve this, we need to find the time it takes for the plane to reach the takeoff speed, and then calculate the distance (runway length) it travels during that time.

Step 1: Find the velocity equation

The velocity is the derivative of the position function with respect to time: $v(t) = \frac{dx}{dt} = \frac{d}{dt}(1.3t^2) = 2 \times 1.3t = 2.6t$

Step 2: Solve for time

We are given the minimum takeoff speed $v = 33.8 \, \text{m/s}$. Set the velocity equal to this speed and solve for $t$: $33.8 = 2.6t$ $t = \frac{33.8}{2.6} = 13 \, \text{seconds}$

Step 3: Calculate the runway length

Now that we know the plane reaches takeoff speed after 13 seconds, substitute this time into the position function $x(t)$ to find the distance traveled (runway length): $x(13) = 1.3 \times (13)^2 = 1.3 \times 169 = 219.7 \, \text{meters}$

Thus, the minimum required runway length is approximately 219.7 meters.

Would you like more details or have any questions about this solution?

Here are some related questions to expand your understanding:

1. How is the velocity function derived from the position function?
2. Why is the derivative used to find velocity in this problem?
3. How would the result change if the takeoff speed was higher or lower?
4. What other factors could affect the minimum runway length in real-life scenarios?
5. Can you think of a situation where this quadratic model for position wouldn't apply?

Tip: Always check units in physics problems to ensure consistency and correctness throughout your calculations.