Question

A 17500 kg jet airplane is flying through some high winds. At some point in time, the airplane is pointing due north, while the wind is blowing from the north and east. If the force on the plane from the jet engines is 36500 N due north, and the force from the wind is 14500 N in a direction 75.0° south of west, what will be the magnitude and direction of the plane's acceleration at that moment? Enter the direction of the acceleration as an angle measured from due west (positive for clockwise, negative for counterclockwise).

Answer 1

Answer:

The magnitude of the acceleration is 1.34 m/s².

The direction of acceleration is 77.9° clockwise with west.

Explanation:

Given that,

Mass of jet = 17500 kg

Force on the plane = 36500 N due north

Force from the wind =14500 N

Angle = 75.0°

We need to calculate the net force on jet plane

Using formula of net force

$F=36500\hat{j}+14500(-\cos70^{\circ}\hat{i}-\sin70^{\circ}\hat{j})$

We need to calculate the acceleration

$a=\dfrac{36500\hat{j}+14500(-\cos70^{\circ}\hat{i}-\sin70^{\circ}\hat{j})}{17500}$

$a=\dfrac{36500 j}{17500}+(\dfrac{-4959.29\ i}{17500}-\dfrac{13625.54\ j}{17500})$

$a=-0.28\ i+1.307\ j$

The magnitude of the acceleration

$a=\sqrt{(0.28)^2+(1.307)^2}$

$a=1.34\ m/s^2$

We need to calculate the direction of the acceleration

Using formula of the direction

$\tan\theta=\dfrac{j}{i}$

Put the value into the formula

$\tan\theta=\dfrac{1.307}{0.28}$

$\theta=\tan^{-1}\dfrac{1.307}{0.28}$

$\theta=77.9^{\circ}$

The direction of acceleration is 77.9° clockwise with west.

Hence, The magnitude of the acceleration is 1.34 m/s².

The direction of acceleration is 77.9° clockwise with west.