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A miniature quadcopter is located at $x_{i}=2.00 \mathrm{m}$ and $y_{i}=$ 4.50 $\mathrm{m}$ at $t=0$ and moves with an average velocity having components $v_{\mathrm{av}, x}=1.50 \mathrm{m} / \mathrm{s}$ and $v_{\mathrm{ax}, y}=-1.00 \mathrm{m} / \mathrm{s}$ . What are the (a) $x$ -coordinate and (b) y-coordinate of the quadcopter's position at $t=2.00 \mathrm{s}$ ?

(a) $x _ { f } = 3 m$

(b) $y _ { f } = - 2 m$

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superpower Today we're going to solve for the, um essentially the final exposition of the quad copters of the corn crop of the quad copter at 2.0 seconds. And so we know that the average velocity in the X direction would be equaling Delta X divided by Delta T. And so this would be equaling X final minus x initial divided by delta t So essentially, X final is equaling ex initial plus the average velocity in the ex direction multiplied by delta T So x final is equaling 2.0 meters plus 1.50 meters per second multiplied by 2.0 seconds and this is giving us 5.0 meters. This would be our final answer for part a and then similarly we can say for part B why final is gonna be equaling. Why initial plus the average velocity in the UAE direction time, Delta t, and this is gonna be equaling 4.50 meters plus negative 1.0 meters per second multiplied by 2.0 seconds and we find that why final is equaling 2.50 meters. This would be our final answer for part B. That is the end of the solution. Thank you for watching