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 ing units. Select one: a. 48 b. 18 c. 88 d. 28 e. 3g

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 />\[{a} = \frac{\Delta \vec{v}}{\Delta t} \]<br /><br />where:<br />- \(\vec{a}\) is the acceleration vector,<br />- \(\Delta \vec{v}\) is the change in velocity vector,<br />- \(\Delta t\) 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 /><br />\[ \Delta \vec{v} = (400i - 300j) - (300i + 400j) \]<br /><br />\[ \Delta \vec{v} = 400i - 300j - 300i - 400j \]<br /><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 /><br />\[ |\Delta \vec{v}| = \sqrt{10000 + 490000} \]<br /><br />\[ |\Delta \vec{v}| = \sqrt{500100} \]<br /><br />\[ |\Delta \vec{v}| = 700\sqrt{10} \]<br /><br />Now, calculate the magnitude of the acceleration:<br /><br />\[ |\vec{a}| = \frac{|\Delta \vec{v}|}{\Delta t} \]<br /><br />\[ |\vec{a}| = \frac{700\sqrt{10}}{10} \]<br /><br />\[ |\vec{a}| = 70\sqrt{10} \]<br /><br />To express this in a more familiar unit (m/s²), we can approximate \(\sqrt{10} \approx 3.16\):<br /><br />\[ |\vec{a}| \approx 70 \times 3.16 \]<br /><br />\[ |\vec{a}| \approx 219.2 \, \text{m/s}^2 \]<br /><br />Therefore, the correct answer is not listed among the provided options. The calculated acceleration is approximately \(219.2 \, \text{m/s}^2\).