# A Cessna aircraft has a liftoff speed of 120 km/h. What minimum constant acceleration does the aircraft require if it is to be airborne after a takeoff run of 240 m?

This article aims to find the acceleration of the aircraft. The article uses the equation of kinematics. Kinematic equations are a set of equations that describe the motion of an object with constant acceleration. Kinematic equations require knowledge of derivatives, rate of change, and integrals. Kinematics equations link five kinematics variables.

1. Displacement $(denoted \: by \: \Delta x)$
2. Initial velocity $(denoted \: by \: v_{o} )$
3. Final velocity $(denoted\: by \: v_{f} )$
4. Time interval $(denoted\: by \: t)$
5. Constant acceleration $(denoted \: by \: a )$

These are basic kinematics equations.

$v = v_ {0} +at$

$v _{f} ^ {2} = v_{i} ^ {2} + 2aS$

$\Delta x = (\dfrac {v + v_{0} }{2} ) t$

## Expert Answer

Aircraft starts from rest. Therefore, the initial speed is:

$v _ {i}= 0.00 \:m s ^ {-1}$

The final speed of the aircraft is:

$v _ {f} = 120\: kmh ^ {-1}$

$= 33.3 \: ms ^ {-1}$

Takeoff run length is:

$\Delta x = 240\: m$

Here, we have the initial velocity, final velocity, and displacement, so we can use the kinematic equation to calculate the acceleration as:

$v _{f} ^ {2} = v_{i} ^ {2} + 2aS$

Rearranging the above equation for acceleration:

$a = \dfrac {v _{f} ^ {2}\: – \:v_{i} ^ {2} } {2S}$

$= \dfrac {(33.3\: m s ^ {-1} ) ^ {2} – (0.00 \: m s ^ {-1}) ^ {2} } {2 \times 240m}$

$= 2.3148 \: m s ^ {-2}$

$a = 2.32 \: m s ^ {-2}$

The acceleration of the aircraft is $2.32 \: m s ^ {-2}$.

## Numerical Result

The acceleration of the aircraft is $2.32 \: m s ^ {-2}$.

## Example

A Cessna airplane has a take-off speed of $150\: \dfrac {km} {h}$. What minimum constant acceleration does the airplane need if it is to be in the air $250\: m$ after takeoff?

Solution

Aircraft starts from rest, therefore the initial speed is:

$v _{i}= 0.00 \: m s ^ {-1}$

The final speed of the aircraft is:

$v_{f} = 150\: kmh ^ {-1}$

$= 41.66 \: ms ^ {-1}$

Takeoff run length is:

$\Delta x = 250 \: m$

Here, we have the initial velocity, final velocity, and displacement, so we can use the kinematic equation to calculate the acceleration as:

$v _{f} ^{2} = v_{i} ^ {2} + 2aS$

Rearranging the above equation for acceleration:

$a = \dfrac {v _ {f} ^ {2}\: – \:v _ {i} ^ {2}} {2S}$

$= \dfrac {(41.66\: m s ^ {-1} ) ^{2} – (0.00 \: m s ^ {-1}) ^ {2} } {2 \times 250m}$

$= 2.47 \: m s ^ {-2}$

$a = 2.47 \: m s ^ {-2}$

The acceleration of the aircraft is $2.47 \: m s ^ {-2}$.

5/5 - (19 votes)