Question

An airplane starts from rest and accelerates at \(12.1 \mathrm{~m} / \mathrm{s}^{2}\). What is its speed at the end of a \(500 .-\mathrm{m}\) runway?

Short answer

Answer: The speed of the airplane at the end of the 500m runway is approximately 110 m/s.

Step-by-step solution

Write down the given values

We know the following values: - Initial velocity (u) = 0 m/s - Acceleration (a) = 12.1 m/s² - Distance (s) = 500 m

Use the equation of motion

Now, we will use the equation v² = u² + 2as to find the final velocity (v).

Substitute the values and solve for v

Substitute the given values into the equation: v² = (0)² + 2(12.1)(500) Calculating the result: v² = 0 + 2(12.1)(500) v² = 12100 Now, find the square root of 12100 to get the final velocity (v): v = √12100 v ≈ 110 m/s

State the final answer

The speed of the airplane at the end of the 500m runway is approximately 110 m/s.