physics
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?
To the problem with vectors, we have to break the trip up into two parts, I'll call them $A$ (airport to change heading) and $B$ (change heading to emergency landing) like you did. From the picture we can see that the vector $A$ uses an angle of $68^{\circ}$ from the vertical. So $A = \langle 140\sin(68^{\circ}), 140\cos(68^{\circ})\rangle \approx \langle 129.806, 52.445\rangle$. You could also have done $A = \langle 140\cos(22^{\circ}), 140\sin(22^{\circ})\rangle$ if you wanted to work with the angle formed between $A$ and the horizontal. For vector $B$, we have the angle with the horizontal but it is below the horizontal. So the vertical component of $B$ is negative. $B = \langle 255\cos(48^{\circ}), -255\sin(48^{\circ})\rangle \approx \langle 170.628, -189.502\rangle$. The resultant is $R = A+B = \langle 300.434, -137.507\rangle$. $|R| = \sqrt{(300.434)^2+(-137.507)^2} \approx 330.220$ km. The angle they need to travel at is $\tan^{-1}\left(\frac{137.507}{300.434}\right) \approx 27.24^{\circ}$ below the horizontal or south of east.
Best Answer
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 $$a=\dfrac{3}{10}g=2.94\;m/s^2.$$