An aircraft has a lift-off speed of 12.1 km/h. If the aircraft's acceleration is constant, what minimum time is required for the aircraft to be airborne after a takeoff run of 23.6 m?
Displacement, Velocity and Acceleration
In classical mechanics, kinematics deals with the motion of a particle. It deals only with the position, velocity, acceleration, and displacement of a particle. It has no concern about the source of motion.
Linear Displacement
The term "displacement" refers to when something shifts away from its original "location," and "linear" refers to a straight line. As a result, “Linear Displacement” can be described as the movement of an object in a straight line along a single axis, for example, from side to side or up and down. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Linear displacement is usually measured in millimeters or inches and may be positive or negative.
![**Question:**
An aircraft has a lift-off speed of 12.1 km/h. If the aircraft's acceleration is constant, what minimum time is required for the aircraft to be airborne after a takeoff run of 23.6 m?
**Explanation:**
This problem requires us to determine the minimum time needed for an aircraft to become airborne, given its lift-off speed and the distance of the takeoff run. To solve this, we can use equations from kinematics that describe motion with constant acceleration.
First, convert the lift-off speed from kilometers per hour to meters per second:
\[ 12.1 \, \text{km/h} = 12.1 \times \frac{1000 \, \text{m}}{3600 \, \text{s}} = 3.36 \, \text{m/s} \]
Next, we will use the kinematic equation that relates distance (s), initial velocity (u), acceleration (a), and time (t):
\[ s = ut + \frac{1}{2}at^2 \]
Since the initial velocity (u) is 0 (the aircraft starts from rest), the equation simplifies to:
\[ s = \frac{1}{2}at^2 \]
Rearranging this equation to solve for the time (t):
\[ t^2 = \frac{2s}{a} \]
\[ t = \sqrt{\frac{2s}{a}} \]
We need the acceleration (a), which we can find using the final velocity (v) and distance (s) with the kinematic equation:
\[ v^2 = u^2 + 2as \]
Again, since the initial velocity (u) is 0:
\[ v^2 = 2as \]
Solving for a:
\[ a = \frac{v^2}{2s} \]
Using the converted velocity (v = 3.36 m/s) and distance (s = 23.6 m):
\[ a = \frac{(3.36 \, \text{m/s})^2}{2 \times 23.6 \, \text{m}} \]
\[ a = \frac{11.29 \, \text{m}^2/\text{s}^2}{47.2 \, \text{m}} \]
\[ a = 0.239 \, \text{m/s](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0f0c4e77-fecd-4d18-90ed-9b329310f28f%2Fe6407c05-f421-49a1-a4c0-2cc6c1e60095%2F1g9o8fn.png&w=3840&q=75)
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