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LUCKY_DIMON [66]
3 years ago
13

A miniature quadcopter is located at x = -2.25 m and y, - 5.70 matt - 0 and moves with an average velocity having components Vv,

x - 2.70 m/s and very - -2.50 m/s. What are the x-coordinate and
y-coordinate (in m) of the quadcopter's position at t= 1.60 ?

(a) x-coordinate
B y-coordinate
Physics
1 answer:
kupik [55]3 years ago
8 0

Recall that average velocity is equal to change in position over a given time interval,

\vec v_{\rm ave} = \dfrac{\Delta \vec r}{\Delta t}

so that the <em>x</em>-component of \vec v_{\rm ave} is

\dfrac{x_2 - (-2.25\,\mathrm m)}{1.60\,\mathrm s} = 2.70\dfrac{\rm m}{\rm s}

and its <em>y</em>-component is

\dfrac{y_2 - 5.70\,\mathrm m}{1.60\,\mathrm s} = -2.50\dfrac{\rm m}{\rm s}

Solve for x_2 and y_2, which are the <em>x</em>- and <em>y</em>-components of the copter's position vector after <em>t</em> = 1.60 s.

x_2 = -2.25\,\mathrm m + \left(2.70\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{x_2 = 2.07\,\mathrm m}

y_2 = 5.70\,\mathrm m + \left(-2.50\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{y_2 = 1.70\,\mathrm m}

Note that I'm reading the given details as

x_1 = -2.25\,\mathrm m \\\\ y_1 = -5.70\,\mathrm m \\\\ v_x = 2.70\dfrac{\rm m}{\rm s}\\\\ v_y=-2.50\dfrac{\rm m}{\rm s}

so if any of these are incorrect, you should make the appropriate adjustments to the work above.

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The coefficients of static and kinetic frictions for plastic on wood are 0.50 and 0.40,respectively. How much horizontal force w
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The magnitude of the force will depend on the coefficient of kinetic friction between the two materials.

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