After a vigorous soccer match, Tina and Michael decide to have a glass of their favorite refreshment. They each run in a straight line along the indicated paths at a speed of 10 ft/sec.
Tina starts at (250, 75) and ends up at (-50, 175)
Michael starts at (0, 0) and ends up at (100, 400)
Write parametric equations for the motion of Tina and Michael individually after t seconds.

Respuesta :

Answer:

The parametric equations for the motion of Tina and Michael are [tex](x, y) = (250-9.487\cdot t, 75+3.162\cdot t)[/tex] and [tex](x,y) = (2.42\cdot t, 9.68\cdot t)[/tex], respectively.

Step-by-step explanation:

Given that both Tina and Michael travels at constant speed in a straight line. We get that vectorial form of the parametric equation of the line is represented by the following expression:

[tex](x, y) = (x_{o}, y_{o})+\frac{v\cdot t}{r} \cdot (x_{f}-x_{o}, y_{f}-y_{o})[/tex] (Eq. 1)

Where:

[tex]x[/tex], [tex]y[/tex] - Current position of the particle, measured in feet.

[tex]x_{o}[/tex], [tex]y_{o}[/tex] - Initial position of the particle, measured in feet.

[tex]x_{f}[/tex], [tex]y_{f}[/tex] - Final position of the particle, measured in feet.

[tex]v[/tex] - Speed of the particle, measured in feet per second.

[tex]t[/tex] - Time, measured in seconds.

[tex]r[/tex] - Distance between initial and final position of the particle, measured in feet.

In addition, the distance between initial and final position of the particle is represented by the following Pythagorean expression:

[tex]r = \sqrt{(x_{f}-x_{o})^{2}+(y_{f}-y_{o})^{2}}[/tex] (Eq. 2)

The parametric equation for each person are, respectively:

Tina

[tex](x_{o},y_{o}) = (250, 75)[/tex], [tex](x_{f}, y_{f}) = (-50, 175)[/tex] and [tex]v = 10\,\frac{ft}{s}[/tex]

From (Eq. 2):

[tex]r = \sqrt{[(-50)-250]^{2}+(175-75)^{2}}[/tex]

[tex]r \approx 316.228\,ft[/tex]

In (Eq. 1):

[tex](x,y) = (250,75)+\frac{2500\cdot t}{79057} \cdot (-300,100)[/tex]

[tex](x,y) = (250, 75) +(-9.487\cdot t, 3.162\cdot t)[/tex]

[tex](x, y) = (250-9.487\cdot t, 75+3.162\cdot t)[/tex]

Michael

[tex](x_{o}, y_{o}) = (0, 0)[/tex], [tex](x_{f}, y_{f})=(100,400)[/tex] and [tex]v = 10\,\frac{ft}{s}[/tex]

From (Eq. 2):

[tex]r = \sqrt{(100-0)^{2}+(400-0)^{2}}[/tex]

[tex]r\approx 412.311\,ft[/tex]

In (Eq. 1):

[tex](x,y) = (0,0)+0.0242\cdot t \cdot (100,400)[/tex]

[tex](x,y) = (2.42\cdot t, 9.68\cdot t)[/tex]

The parametric equations for the motion of Tina and Michael are [tex](x, y) = (250-9.487\cdot t, 75+3.162\cdot t)[/tex] and [tex](x,y) = (2.42\cdot t, 9.68\cdot t)[/tex], respectively.

The parametric equation that can be used to determine Tina's and

Michael's location are expressed as follows;

  • Tina's parametric equation is; [tex]\underline{(x, \, y) = (250 - 9.49 \cdot t, \ 75 + 3.16\cdot t)}[/tex]  
  • Michael's parametric equation is; [tex]\underline{(x, \, y) = (2.425 \cdot t, \ 9.7\cdot t)}[/tex]

Reasons:

The given parameters are;

The speed of the runners = 10 ft./sec

Location Tina starts = (250, 75)

Point Tina ends = (-50, 175)

Location Michael starts = (0, 0)

Point Michael ends = (100, 400)

Parametric equation for Tina's motion;

[tex]The \ \mathbf{slope} \ of \ Tina's \ path \ is \ \dfrac{175 - 75}{-50 - 250} = -\dfrac{100}{300} = -\dfrac{1}{3}[/tex]

[tex]The \ \mathbf{direction} \ is \ therefore; \ arctan \left( -\dfrac{1}{3} \right) \approx 161.57^{\circ}[/tex]

[tex]The \ speed \ in \ the \ x-direction \ is \ v_x = 10 \times cos \left(arctan \left( -\dfrac{1}{3} \right)\right) \approx - 9.49[/tex]

Therefore;

The location in the x-direction after t seconds is x ≈ 250 - 9.49·t

[tex]The \ speed \ in \ the \ x-direction \ is \ v_y = 10 \times sin \left(arctan \left( -\dfrac{1}{3} \right)\right) \approx 3.16[/tex]

The location in the y-direction after t seconds is y ≈ 75 + 3.16·t

The parametric equation for the location of Tina after t seconds is

therefore;

  • [tex]\underline{(x, \, y) = \mathbf{(250 - 9.49 \cdot t, \ 75 + 3.16\cdot t)}}[/tex].

Parametric equation for Michael's motion;

[tex]The \ direction \ of \ Michael's \ motion \ is \ arctan \left( \dfrac{400}{100} \right) \approx 75.96^{\circ}[/tex]

Therefore;

[tex]v_x = 10 \times cos \left(arctan \left(4 \right)\right) \approx 2.425[/tex]

x ≈ 0 + 2.425·t = 2.425·t

[tex]v_y = 10 \times sin \left(arctan \left( 4\right)\right) \approx 9.7[/tex]

y ≈ 9.7·t

The parametric equation for the location of Michael at time t is therefore;

  • [tex]\underline{(x, \, y) = \mathbf{(2.425 \cdot t, \ 9.7\cdot t)}}[/tex]

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