The given data,

\begin{eqnarray*}

\text{Initial velocity},\,v_0 &=& 0\\

\text{Acceleration},\,a &=& 1.9\,{\rm m/s^{2}} \\

\text{Time interval},\,t &=& 5\,{\rm s} \\

\end{eqnarray*}

(a) Use the following equation,

\begin{eqnarray*}

x-x_0 &=& \frac{1}{2}\,at^{2}+v_0 t \\

x- 0 &=& \frac{1}{2}\,(1.9)(5)^{2}+0(5) \\

x &=& 23.75\,{\rm m/s}

\end{eqnarray*}

In the second line, for convenience, we simply adopt the initial position ($x_0$) at time $t=0$ as $0$.

(b) the equation below gives the velocity at the end of time interval

\begin{eqnarray*}

v &=& v_0+at \\

v &=& 0+(1.9)(5)\\

&=& +9.5\,{\rm m/s} \qquad {West}

\end{eqnarray*}

(c) In a straight line motion, if the velocity and acceleration have the same signs, the speed of the moving object increases. Here, by establishing a coordinate system and choosing west as positive direction, we can see the acceleration and velocity are in the same direction. Therefore, the object moves west without any changing direction. In this type of motions, which object does not change its direction, the displacement (vector) and distance traveled is the same. Thus, as calculated in (a), the total distance is approximately $24\,{\rm m}$.

(d) As reasoning of (c), since the direction of the motion does not changes so the magnitude of vector quantities shows the value of scalar ones. Here, the magnitude of final velocity ($v=9.5\,{\rm m/s}$) is equal to the final speed.