An aircraft makes a horizontal circular turn. The air craft enters the turn with velocity of v=300i+400j 10 seconds later it leaves the turn with velocity of v=400i-300j Determine the acceleration of the aircraft in g units. Select one: a. 48 b. 18 c. 88 d. 28 e. 38

To determine the acceleration of the aircraft, we need to calculate the change in velocity over time. The acceleration vector is given by:<br /><br />\[ \vec{a} = \frac{\Delta \vec{v}}{\Delta t} \]<br /><br />where \( \Delta \vec{v} \) is the change in velocity and \( \Delta t \) is the change in time.<br /><br />Given:<br />Initial velocity, \( \vec{v}_i = 300i + 400j \)<br />Final velocity, \( \vec{v}_f = 400i - 300j \)<br />Time interval, \( \Delta t = 10 \) seconds<br /><br />First, calculate the change in velocity:<br /><br />\[ \Delta \vec{v} = \vec{v}_f - \vec{v}_i \]<br />\[ \Delta \vec{v} = (400i - 300j) - (300i + 400j) \]<br />\[ \Delta \vec{v} = 400i - 300j - 300i - 400j \]<br />\[ \Delta \vec{v} = 100i - 700j \]<br /><br />Next, calculate the magnitude of the change in velocity:<br /><br />\[ |\Delta \vec{v}| = \sqrt{(100)^2 + (-700)^2} \]<br />\[ |\Delta \vec{v}| = \sqrt{10000 + 490000} \]<br />\[ |\Delta \vec{v}| = \sqrt{500100} \]<br />\[ |\Delta \vec{v}| \approx 707.72 \]<br /><br />Now, calculate the magnitude of the acceleration:<br /><br />\[ |\vec{a}| = \frac{|\Delta \vec{v}|}{\Delta t} \]<br />\[ |\vec{a}| = \frac{707.72}{10} \]<br />\[ |\vec{a}| \approx 70.77 \]<br /><br />Finally, convert the acceleration to g units (where 1 g ≈ 9.81 m/s²):<br /><br />\[ |\vec{a}|_{\text{g units}} = \frac{70.77}{9.81} \]<br />\[ |\vec{a}|_{\text{g units}} \approx 7.21 \]<br /><br />Therefore, the correct answer is not listed among the options provided. However, based on the closest value, the correct answer should be approximately 7.21 g units.