The average speed equation is

$$
begin{align*}
text{average speed}=dfrac{text{distance}}{text{elapsed time}}\
end{align*}
$$

Rearranging the equation to solve for distance and plugging in the values, we obtain the result:

$$
begin{align*}
text{distance}&=text{average speed}timestext{elapsed time}\
&=4.6 frac{text{m}}{text{s}}times 1.4 text{s}\
&=quadboxed{6.4 text{m}}\
end{align*}
$$

$$
begin{align*}
boxed{text{distance}=6.4 text{m}}\
end{align*}
$$

$text{underline{Answer in meters}}$

Using the equation for average speed, rearranging it to solve for distance, and plugging in the values, we get

$$
begin{align*}
text{average speed}&=dfrac{text{distance}}{text{elapsed time}}\
text{distance}&=text{average speed}timestext{elapsed time}\
&=340 frac{text{m}}{text{s}}times 3.5 text{s}\
&=quadboxed{1190 text{m}}\
end{align*}
$$

$text{underline{Answer in kilometers}}$

The distance in kilometers is simply the distance in meters divided by 1000, as a meter is one thousandth of a kilometer,

$$
begin{align*}
boxed{text{distance}=1.190 text{km}}\
end{align*}
$$

If you want a more detailed approach for the answer in kilometers, you can convert $frac{text{m}}{text{s}}$ to $frac{text{km}}{text{h}}$ and seconds to hours.

To convert $frac{text{m}}{text{s}}$ to $frac{text{km}}{text{h}}$, multiply the speed value by a conversion factor $left(dfrac{3.6 frac{text{km}}{text{h}}}{1 frac{text{m}}{text{s}}}right)$, and to convert seconds to hours, multiply the time value by $left(dfrac{1 text{h}}{3600 text{s}}right)$. Therefore,

$$
begin{align*}
text{distance}&=340 frac{text{m}}{text{s}}cdotleft(dfrac{3.6 frac{text{km}}{text{h}}}{1 frac{text{m}}{text{s}}}right)times 3.5 text{s}cdotleft(dfrac{1 text{h}}{3600 text{s}}right)\
&=1224 frac{text{km}}{text{h}}timesfrac{3.5 text{h}}{3600}\
&=quadboxed{1.19 text{km}}\
end{align*}
$$

In both cases, the final result is the same, but the first approach is more simple and straightforward.

$$
begin{align*}
text{In kilometers:} quad &boxed{text{distance}=1190 text{m}}\
text{In meters:} quad &boxed{text{distance}=1.19 text{km}}\
end{align*}
$$

$textbf{(a)}$ We will again use the equation for average speed and solve it for time. However, we have to convert minutes to hours to get a dimensionally consistent result.

$$
begin{align*}
text{average speed}&=dfrac{text{distance}}{text{elapsed time}}\
text{distance}&=text{average speed}timestext{elapsed time}\
&=65 frac{text{km}}{text{h}}times 3.2 text{min}cdotleft(dfrac{1 text{h}}{60 text{min}}right)\
&=65 frac{text{km}}{text{h}}timesdfrac{3.2 text{h}}{60}\
&=quadboxed{3.5 text{km}}\
end{align*}
$$

$textbf{(b)}$ Solving the same equation for the elapsed time and plugging in the values, we get

$$
begin{align*}
text{average speed}&=dfrac{text{distance}}{text{elapsed time}}\
text{elapsed time}&=dfrac{text{distance}}{text{average speed}}\
&=dfrac{0.25 text{km}}{65 frac{text{km}}{text{h}}}\
&=quadboxed{0.0038 text{h}}\
end{align*}
$$

When dealing with such small numbers, it’s more intuitive and makes more sense to write the result in appropriate order-of-magnitude units. For that reason, we’ll write the solution for elapsed time in seconds:

$$
begin{align*}
boxed{text{elapsed time}=14 text{s}}\
end{align*}
$$