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

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

Answer 2

The coordinates X and Y at t = 1.60 seconds of the quadcopter are 4.32 m and - 9.70 m respectively.

Recall :

Distance = Speed × time

X - coordinate:

• Initial position at t = 0 ; $X_i = - 1.75 m$
• Average velocity = 2.70 m/s

• At t = 1.60 s
• Distance moved = 2.70 m/s × 1.60 s = 4.32 m

• Distance moved for t = 1.60 s
• Initial position + distance moved
• -1.75 + 4.32 = 2.57 m

Y - coordinate :

• Initial position at t = 0 ; $Y_i = - 5.70 m$
• Average velocity = -2.50 m/s

• Distance moved for t = 1.60 s
• Distance moved = - 2.50m/s × 1.60 s = - 4.00 m

• Distance moved for t= 1.60 s
• Initial position + distance moved
• -5.70 + (-4.00) = - 9.70 m

Therefore, the x and y coordinate of the quadcopter are : 2.57 m and - 9.70 m respectively.

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