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?

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**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
Transcribed Image Text:**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
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