Question

6. An airplane of mass 1.5 x 104 kg is moving at 60 m/s. The pilot then revs the engine so that the forward thrust by the engines becomes 7.5 x 104 N. If the force of air resistance is 4.0 x 104 N , find the speed of the airplane after it has traveled 800 m. Assume that the plane remains horizontal throughout this flight.

Answer 1

Answer:

The speed of the airplane after it has travelled 800 meters is approximately 85.632 meters per second.

Explanation:

From statement, we obtain the equation of equilibrium associated with the airplane:

$\Sigma F = T - R = m\cdot a$ (1)

Where:

$T$ - Thrust force from engines, measured in newtons.

$R$ - Force of air resistance, measured in newtons.

$m$ - Mass of the airplane, measured in kilograms.

$a$ - Net acceleration of the airplane, measured in meters per square second.

If we know that $T = 7.5\times 10^{4}\,N$, $R = 4.0\times 10^{4}\,N$ and $m = 1.5\times 10^{4}\,kg$, then the net acceleration of the airplane is:

$a = \frac{T-R}{m}$

$a = \frac{7.5\times 10^{4}\,N-4.0\times 10^{4}\,N}{1.5\times 10^{4}\,kg}$

$a = 2.333\,\frac{m}{s^{2}}$

Lastly, the final speed of the airplane is calculated by the following kinematic equation:

$v_{f} = \sqrt{v_{o}^{2}+2\cdot a \cdot \Delta s}$ (2)

Where:

$\Delta s$ - Travelled distance, measured in meters.

$v_{o}$, $v_{f}$ - Initial and final speeds of the aircraft, measured in meters per second.

If we know that $v_{o} = 60\,\frac{m}{s}$, $a = 2.333\,\frac{m}{s^{2}}$ and $\Delta s = 800\,m$, then the final speed of the aircraft is:

$v_{f} = \sqrt{\left(60\,\frac{m}{s} \right)^{2}+2\cdot \left(2.333\,\frac{m}{s^{2}} \right)\cdot (800\,m)}$

$v_{f} \approx 85.632\,\frac{m}{s}$

The speed of the airplane after it has travelled 800 meters is approximately 85.632 meters per second.