To determine the minimum acceleration required for the airplane to reach the takeoff speed of 75 m/s over a runway length of 950 meters, we can use the kinematic equation:

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

where:

• $v = 75 \, \text{m/s}$ is the final velocity (takeoff speed),
• $u = 0 \, \text{m/s}$ is the initial velocity (since it starts from rest),
• $a$ is the acceleration, and
• $s = 950 \, \text{m}$ is the distance (runway length).

Rearranging to solve for $a$:

$a = \frac{v^2 - u^2}{2s}$

Substitute the values:

$a = \frac{(75)^2 - (0)^2}{2 \times 950}$

$a = \frac{5625}{1900}$

$a \approx 2.96 \, \text{m/s}^2$

Answer:

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

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

Here are some related questions for practice:

1. What would the required acceleration be if the runway length was only 800 m?
2. How long would it take the airplane to reach takeoff speed under this acceleration?
3. If the initial speed wasn’t zero but 10 m/s, what would be the required acceleration?
4. What runway length would be needed if the maximum possible acceleration was only $2 \, \text{m/s}^2$?
5. How would air resistance impact the actual acceleration needed?

Tip: Always keep track of initial and final velocities, as well as displacement, when applying kinematic equations.