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:
We'll apply this to both the initial speed (22 knots) and the final speed (153 knots).
Initial speed (22 knots):
Final speed (153 knots):
Step 2: Use the kinematic equation to find the runway distance
We can use the following kinematic equation to find the distance (in feet):
Where:
- is the final velocity (258.1875 ft/sec),
- is the initial velocity (37.125 ft/sec),
- is the acceleration (9.8 ft/sec²),
- is the distance (runway used, in feet).
Rearranging the equation to solve for :
Substitute the known values:
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:
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:
- How would the total runway distance change if the acceleration were increased to 12 feet/sec²?
- What other kinematic equations could be used to solve this problem?
- How do the units convert between knots and feet per second in more detail?
- How does the final velocity relate to the required takeoff speed for this aircraft?
- 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
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