4. A particle moves in the $$ \mathrm{x}-\mathrm{y} $$ plane, starting from the origin at $$ \mathrm{t}=0 $$ with an initial velocity having an x -component of $$ -20 \mathrm{~m} / \mathrm{s} $$ and a y -component of $$ -15 \mathrm{~m} / \mathrm{s} $$, as shown in the figure below. The particle experiences an acceleration in the x -direction, given by $$ a_{x}= $$ $$ 2.0 \mathrm{~m} / \mathrm{s}^{2} $$. a. Determine the x and y coordinates of the particle at any time, $$ \mathrm{t}=5 \mathrm{~s} $$, and its position vector at that time. b. Determine the velocity and speed of the particle at $$ \mathrm{t}=5.0 \mathrm{~s} $$. c. Determine the angle the velocity vector makes with the x axis.

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**a. Position at $$ t=5.0\,\mathrm{s} $$**

Since the acceleration is only in the x-direction, we have:  
- For the x-coordinate:  
  $$
  x = x_0 + u_x t + \frac{1}{2}a_x t^2
  $$
  Given $$ x_0 = 0 $$, $$ u_x = -20\,\mathrm{m/s} $$, $$ a_x = 2.0\,\mathrm{m/s}^2 $$, and $$ t = 5.0\,\mathrm{s} $$:
  $$
  x = 0 + (-20)(5.0) + \frac{1}{2}(2.0)(5.0^2) = -100 + \frac{1}{2}(2.0)(25) = -100 + 25 = -75\,\mathrm{m}
  $$

- For the y-coordinate:  
  There is no acceleration in the y-direction, so:
  $$
  y = y_0 + u_y t
  $$
  Given $$ y_0 = 0 $$ and $$ u_y = -15\,\mathrm{m/s} $$:
  $$
  y = 0 + (-15)(5.0) = -75\,\mathrm{m}
  $$

Thus, the position vector at $$ t=5.0\,\mathrm{s} $$ is:
$$
\vec{r} = \langle -75, \ -75 \rangle\,\mathrm{m}
$$

---

**b. Velocity and Speed at $$ t=5.0\,\mathrm{s} $$**

- For the x-component of velocity:
  $$
  v_x = u_x + a_x t = -20 + 2.0(5.0) = -20 + 10 = -10\,\mathrm{m/s}
  $$

- For the y-component of velocity (constant):
  $$
  v_y = u_y = -15\,\mathrm{m/s}
  $$

The velocity vector is:
$$
\vec{v} = \langle -10, \ -15 \rangle\,\mathrm{m/s}
$$

The speed (magnitude of velocity) is:
$$
|\vec{v}| = \sqrt{v_x^2 + v_y^2} = \sqrt{(-10)^2 + (-15)^2} = \sqrt{100 + 225} = \sqrt{325} = 5\sqrt{13}\,\mathrm{m/s}
$$

---

**c. Angle the Velocity Vector Makes with the x-axis**

The angle $$ \theta $$ (measured from the positive x-axis) is given by:
$$
\tan \theta = \frac{v_y}{v_x} = \frac{-15}{-10} = 1.5
$$
Calculating the reference angle:
$$
\theta_{\text{ref}} = \arctan(1.5) \approx 56.3^\circ
$$

Since both $$ v_x $$ and $$ v_y $$ are negative, the velocity vector is in the third quadrant. Therefore, the actual angle measured from the positive x-axis is:
$$
\theta = 180^\circ + 56.3^\circ \approx 236.3^\circ
$$

Alternatively, one may report the angle relative to the negative x-axis as approximately $$ 56.3^\circ $$ (below the negative x-direction).

---
**a. Position at $$ t=5.0\,\mathrm{s} $$:** - $$ x = -75\,\mathrm{m} $$ - $$ y = -75\,\mathrm{m} $$ - Position vector: $$ \vec{r} = \langle -75, \ -75 \rangle\,\mathrm{m} $$ **b. Velocity and Speed at $$ t=5.0\,\mathrm{s} $$:** - Velocity vector: $$ \vec{v} = \langle -10, \ -15 \rangle\,\mathrm{m/s} $$ - Speed: $$ 5\sqrt{13}\,\mathrm{m/s} $$ (approximately $$ 22.9\,\mathrm{m/s} $$) **c. Angle the Velocity Vector Makes with the x-axis:** - Approximately $$ 236.3^\circ $$ (or $$ 56.3^\circ $$ below the negative x-axis)

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