Question

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?

Answer 1

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

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