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**.

**Displacement**$(denoted \: by \: \Delta x)$**Initial velocity**$(denoted \: by \: v_{o} )$**Final velocity**$ (denoted\: by \: v_{f} )$**Time interval**$ (denoted\: by \: t) $**Constant acceleration**$ (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} $.