To solve for the minimum acceleration required for the airplane to reach the takeoff speed, we can use the kinematic equation:

$v^2 = u^2 + 2as$

where:

• $v$ is the final velocity (takeoff speed) = 75 m/s,
• $u$ is the initial velocity = 0 (since the plane starts from rest),
• $a$ is the acceleration (what we need to find),
• $s$ is the distance (runway length) = 950 m.

Plugging in the known values:

$(75)^2 = 0 + 2 \times a \times 950$

Solving for $a$:

$5625 = 1900a$

$a = \frac{5625}{1900} = 2.96 \, \text{m/s}^2$

Answer:

The minimum acceleration required is approximately $2.96 \, \text{m/s}^2$.

Would you like a detailed explanation on each step or have any questions?

Here are 5 related questions to consider:

1. How would the required acceleration change if the runway length were doubled?
2. What minimum runway length is required if the plane’s acceleration capability is limited to 2 m/s²?
3. How does the minimum takeoff speed of an airplane relate to its weight and design?
4. If the airplane had an initial velocity of 10 m/s, what would the required acceleration be?
5. How does air resistance affect takeoff requirements in real scenarios?

Tip: Always double-check your units when working with physics equations to avoid errors in calculations.