Math Problem Statement

An aircraft has an acceleration of 9 . 8 feet / sec ^ 2 .

From a rolling start of 2 2 knots and 1 5 6 feet down the runway, how much total runway, in feet, is used to reach a Vr of 1 5 3 knots? Round your answer to 1 decimal place.

Solution

To solve this problem, we need to calculate the total runway distance used by the aircraft to reach the desired speed.

Step 1: Convert the speeds from knots to feet per second

1 knot = 1 nautical mile per hour = 1.15078 miles per hour 1 mile = 5280 feet, so:

Speed in feet per second=Speed in knots×1.15078×52803600\text{Speed in feet per second} = \text{Speed in knots} \times 1.15078 \times \frac{5280}{3600}

We'll apply this to both the initial speed (22 knots) and the final speed (153 knots).

Initial speed (22 knots):

22knots=22×1.15078×52803600=22×1.6875=37.125ft/sec22 \, \text{knots} = 22 \times 1.15078 \times \frac{5280}{3600} = 22 \times 1.6875 = 37.125 \, \text{ft/sec}

Final speed (153 knots):

153knots=153×1.15078×52803600=153×1.6875=258.1875ft/sec153 \, \text{knots} = 153 \times 1.15078 \times \frac{5280}{3600} = 153 \times 1.6875 = 258.1875 \, \text{ft/sec}

Step 2: Use the kinematic equation to find the runway distance

We can use the following kinematic equation to find the distance dd (in feet):

vf2=vi2+2adv_f^2 = v_i^2 + 2ad

Where:

  • vfv_f is the final velocity (258.1875 ft/sec),
  • viv_i is the initial velocity (37.125 ft/sec),
  • aa is the acceleration (9.8 ft/sec²),
  • dd is the distance (runway used, in feet).

Rearranging the equation to solve for dd:

d=vf2vi22ad = \frac{v_f^2 - v_i^2}{2a}

Substitute the known values:

d=(258.1875237.1252)2×9.8d = \frac{(258.1875^2 - 37.125^2)}{2 \times 9.8}

d=66768.61376.619.6d = \frac{66768.6 - 1376.6}{19.6}

d=6539219.63330.3feetd = \frac{65392}{19.6} \approx 3330.3 \, \text{feet}

Step 3: Add the initial runway distance

We are also told that the aircraft has already traveled 156 feet down the runway. So, the total runway distance is:

Total runway distance=3330.3ft+156ft=3486.3ft\text{Total runway distance} = 3330.3 \, \text{ft} + 156 \, \text{ft} = 3486.3 \, \text{ft}

Final Answer:

The total runway distance used is approximately 3486.3 feet.

Would you like to go over any part of this calculation in more detail?

Additional questions:

  1. How would the total runway distance change if the acceleration were increased to 12 feet/sec²?
  2. What other kinematic equations could be used to solve this problem?
  3. How do the units convert between knots and feet per second in more detail?
  4. How does the final velocity relate to the required takeoff speed for this aircraft?
  5. What factors could affect the acceleration during the aircraft's takeoff?

Tip: Always double-check your unit conversions before plugging them into equations, especially when dealing with velocity or distance.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Unit Conversion
Quadratic Equations

Formulas

Speed in feet per second = Speed in knots × 1.15078 × (5280/3600)
Kinematic equation: v_f^2 = v_i^2 + 2ad
Runway distance: d = (v_f^2 - v_i^2) / (2a)

Theorems

Kinematic equations of motion

Suitable Grade Level

Grades 10-12