The problem states: You are sitting in an airplane when it encounters sudden turbulence. During the turbulence, you feel lighter for a few moments. If your apparent weight during this time seems to be about $70\%$ of your normal weight, what are the magnitude and direction of the plane’s acceleration?
The answer is $2.94$ $m/s^2$.
How do i go about approaching this problem and setting it up?
On
There is a upward force $\bf N$ which is the normal force (your apparent weight); there is a downward force $m\mathbf{g}$ ($m$ is your mass and $\bf g$ is the gravitational acceleration) which is your weight and there is a total acceleration $\bf a$ that you experiment (this acceleration is in fact the airplane's acceleration) which is directed downwards (because you feel lighter so the plane must be accelerating downwards). Then, by Newton's second law $$m\mathbf{g}-\mathbf{N}=m\mathbf{a}$$ and since all forces occur in the same direction, we can only focus on magnitudes: $$\tag{1}mg-N=ma.$$ Since you know that $N=70\%\;mg=7mg/10$ then $$mg-\dfrac{7}{10}mg=\dfrac{3}{10}mg=ma$$ and since $m\neq0,$ then $$a=\dfrac{3}{10}g=2.94\;m/s^2.$$
On
The other two answers show how to determine the vertical component of the acceleration due to the turbulence if the aircraft is in level flight and not turning. Even in that case, the horizontal component of the acceleration is undetermined: you don't have enough information to calculate the direction of the acceleration.
Your normal weight you feel is $$g\times m $$ where $m$ is your mass in kg's and $g=9.81 \frac{\text{m}}{\text{s}^2},$ the gravitational acceleration at which you would accelerate toward the earth if the floor would not hold you.
If the plane accelerates downward by $a$ then (since your mass does not change) your weight you feel becomes
$$(g-a) \times m .$$ Here $g-a$ would be your acceleration relative to the plain if its floor would not hold you. Regarding your weight feeling $g-a$ plays the same role as $g$ would under normal circumstances.
If $(g-a) \times m $ is $70\%$ of your normal weight then you have the following equation
$$(g-a) \times m =0.7\times g\times m .$$ $m$ cancels out, that is,
$$(g-a)=0.7\times g$$
from where
$$a=0.3\times g=0.3\times 9.81=2.943 \ \frac{\text{m}}{\text{s}^2}.$$