A plane accelerates from rest at a constant rate of 6.00  m/s2
  along a runway that is 1860  m
  long. Assume that the plane reaches the required takeoff velocity at the end of the runway.What is the distance
  traveled by the plane in the last second before taking off?

To solve this problem, we will use the equations of motion under constant acceleration to find the distance traveled by the plane in the last second before takeoff.

### Step-by-Step Solution

1. **Determine the takeoff velocity:**

   Given:
   - Initial velocity \( u = 0 \) (since the plane starts from rest)
   - Constant acceleration \( a = 6.00 \, \text{m/s}^2 \)
   - Runway length \( s = 1860 \, \text{m} \)

   We can use the following equation of motion to find the final velocity \( v \) (takeoff velocity) at the end of the runway:

   \[
   v^2 = u^2 + 2as
   \]

   Substituting the values:

   \[
   v^2 = 0 + 2 \times 6.00 \times 1860
   \]

   \[
   v^2 = 22320
   \]

   \[
   v = \sqrt{22320}
   \]

   Let's calculate the value of \( v \).

2. **Calculate the time to reach takeoff velocity:**

   Now that we have \( v \), we can find the time \( t \) it takes for the plane to reach the takeoff velocity using the formula:

   \[
   v = u + at
   \]

   Substituting the values:

   \[
   t = \frac{v - u}{a}
   \]

   Let's compute the value of \( t \).

3. **Determine the distance traveled in the last second:**

   The total distance traveled by the plane is given by:

   \[
   s(t) = ut + \frac{1}{2}at^2
   \]

   Since the plane starts from rest, \( u = 0 \):

   \[
   s(t) = \frac{1}{2}at^2
   \]

   The distance traveled in the last second before takeoff is the difference between the total distance traveled at time \( t \) and the distance traveled at time \( t - 1 \):

   \[
   \Delta s = s(t) - s(t-1)
   \]

   Substituting:

   \[
   \Delta s = \frac{1}{2}a t^2 - \frac{1}{2}a (t-1)^2
   \]

   Simplify the expression to find \( \Delta s \).

Now, let's perform the calculations.1. **Takeoff velocity**: The plane reaches a velocity of approximately \(149.40 \, \text{m/s}\) at the end of the runway.

2. **Time to reach takeoff velocity**: The time required to reach this velocity is approximately \(24.90 \, \text{seconds}\).

3. **Distance traveled in the last second before takeoff**: The plane travels approximately \(146.40 \, \text{meters}\) in the last second before taking off.

Would you like more details or have any questions? Here are some related questions you might find interesting:

1. What would the takeoff velocity be if the runway length were different?
2. How does changing the acceleration affect the time and distance calculations?
3. What is the significance of the last second's distance in determining takeoff parameters?
4. How can we calculate the force required for the plane to achieve this acceleration?
5. How would the plane's mass affect its acceleration and takeoff velocity?

**Tip:** Knowing how to calculate distances under constant acceleration is essential for understanding various physical scenarios, like car braking distances or objects in free fall.

A plane accelerates from rest at a constant rate of 6.00 m/s^2 along a runway that is 1860 m long. Assume that the plane reaches the required takeoff velocity at the end of the runway. What is the distance traveled by the plane in the last second before taking off?

Learn how to calculate the distance traveled by a plane accelerating at 6.00 m/s^2 along a 1860 m runway in the last second before takeoff. This problem applies the equations of motion under constant acceleration, using key formulas and step-by-step solutions. Suitable for students in Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation