Question

What is the magnitude of the acceleration, in g units, of a pilot whose aircraft enters a horizontal circular turn with a

velocity of vi = ሺ400i + 500j ሻ m/s and 24.0 s later leaves the turn with a velocity of vf = ሺ−400i − 500jሻ m/s?

Answer 1

$$a=\dfrac{v_f-v_i}{t},$$ $$a=\dfrac{(-400i-500j-400i-500j)\ \dfrac{m}{s}}{24\ s},$$ $$a=\dfrac{(-800i-1000j)\ \dfrac{m}{s}}{24\ s}=(-33.33i-41.67j)\ \dfrac{m}{s^2},$$ $$a=\sqrt{a_x^2+a_y^2}=\sqrt{(-33.33\ \dfrac{m}{s^2})^2+(-41.67\ \dfrac{m}{s^2})^2},$$ $$a=53.36\ \dfrac{m}{s^2}\cdot\dfrac{1\ g}{9.8\ \dfrac{m}{s^2}}=5.44g.$$