The minimum takeoff speed for a certain airplane in 75 m/s. What minimum acceleration is requied if the plane must leave a runway of length 950 m ? Assume the plane starts from rest at one end of the runway .

Options

A. `2.0 m//s^(2)`, upward
B. `5.3 m//s^(2)`, upward
C. `2.0 m//s^(2)` , downward
D. `5.3 m//s^(2)`, downward

Answer

To solve the problem of finding the minimum acceleration required for an airplane to take off from a runway of length 950 m, given that its minimum takeoff speed is 75 m/s and it starts from rest, we can use the equations of motion.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial velocity (u) = 0 m/s (the plane starts from rest)
- Final velocity (v) = 75 m/s (minimum takeoff speed)
- Displacement (s) = 950 m (length of the runway)

2. **Use the Equation of Motion:**
We can use the third equation of motion, which relates initial velocity, final velocity, acceleration, and displacement:
\[ v^2 = u^2 + 2as \]
where:
- \( v \) = final velocity
- \( u \) = initial velocity
- \( a \) = acceleration
- \( s \) = displacement

3. **Substitute the Known Values:**
Plugging in the known values into the equation:
\[ (75)^2 = (0)^2 + 2a(950) \]

4. **Calculate \( v^2 \):**
\[ 75^2 = 5625 \]
So, the equation becomes:
\[ 5625 = 2a(950) \]

5. **Simplify the Equation:**
\[ 5625 = 1900a \]

6. **Solve for Acceleration (a):**
\[ a = \frac{5625}{1900} \]
\[ a \approx 2.9684 \, \text{m/s}^2 \]

7. **Round the Result:**
The minimum acceleration required is approximately:
\[ a \approx 2.97 \, \text{m/s}^2 \]
This can be rounded to 3 m/s² for practical purposes.

### Final Answer:
The minimum acceleration required for the airplane to take off is approximately **3 m/s²**.

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