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marissa [1.9K]
3 years ago
6

A miniature quadcopter is located at xi = −1.75 m and yi = −5.70 m at t = 0 and moves with an average velocity having components

vav, x = 2.70 m/s and vav, y = −2.50 m/s. What are the x-coordinate and y-coordinate (in m) of the quadcopter's position at t = 1.60 s?
Mathematics
1 answer:
Black_prince [1.1K]3 years ago
4 0

The quadcopter's average velocity is given in terms of change in position by

\vec v_{\rm av}=\dfrac{\Delta x}{\Delta t}\,\vec\imath+\dfrac{\Delta y}{\Delta t}\,\vec\jmath

where \Delta x=x_f-x_i, the difference in the quadcopter's final and initial positions and \Delta t=(1.60-0)\,\mathrm s=1.60\,\mathrm s.

The x-component of the average velocity is 2.70 m/s, so

2.70\dfrac{\rm m}{\rm s}=\dfrac{x_f-(-1.75\,\mathrm m)}{1.60\,\rm s}\implies \boxed{x_f=2.57\,\mathrm m}

and the y-component is -2.50 m/s, so

-2.50\dfrac{\rm m}{\rm s}=\dfrac{y_f-(-5.70\,\mathrm m)}{1.60\,\rm s}\implies\boxed{y_f=-9.70\,\mathrm m}

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