Answer
$\textbf{(a)}$ $x(5\text{ s})=-55\text{ m},\quad y(5\text{ s})=31\text{ m}.$
$\textbf{(b)}$ $v_x(5\text{ s})=-22\text{ m/s};\quad v_y(5\text{ s})=6.2\text{ m/s}$
$\textbf{(c)}$ The speed only increases with time.
Work Step by Step
The position law of the uniformly accelerating particle is
$x(t)=x_0+v_{x0} t+\frac{1}{2}a_x t^2;\\
y(t)=y_0+v_{y0} t + \frac{1}{2}a_y t^2.$
The velocity laws are:
$v_x(t)=v_{x0}+a_xt;\\
v_y(t)=v_{y0}+a_yt.$
where $x_0$ and $y_0$ are initial $x$ and $y$ coordinates and $v_{x0}$, $v_{y0}$, $a_x$ and $a_y$ initial velocity and acceleraton projetions onto $x$ and $y$ directions, respectively. From the problem we see that: $x_0=y_0=0$ (we choose to measure time when the particle passes through the origin). Further, $v_{x0}=0$, $v_{y0}=6.2\text{ m/s}$, $a_x=-4.4\text{ m/s}^2$ and $a_y=0$.
$\textbf{(a)}$ Using the given data we get
$$x(5\text{ s})=0+0\cdot 5\text{ s}+\frac{1}{2}(-4.4\text{ m/s}^2)\cdot (5\text{ s})^2=-55\text{ m};\\
y(5\text{ s})=0+6.2\text{ m/s}\cdot 5\text{ s}+\frac{1}{2}\cdot 0\cdot (5\text{ s})^2=31\text{ m}.$$
$\textbf{(b)}$ Using the given data we get
$$v_x(5\text{ s})=0+(-4.4\text{ m/s}^2)\cdot5\text{ s}=-22\text{ m/s};\\
v_y(5\text{ s})=6.2\text{ m/s}+0\cdot 5\text{ s}=6.2\text{ m/s}.$$
$\textbf{(c)}$ The speed, which is interpretes as the intensity of the velocity is given by
$$v(t)=\sqrt{v_x(t)^2+v_y(t)^2}.$$
From the given data, we see that $v_y$ is constant, while $v_x$ increases in magnitude (it becomes 'more and more negative'). This means that the argument of the square root only increases and thus, $v$ increases with $t$.
