varies directly with the time Since Tiffaniqua's speed is the constant of variation is The distance traveled
. Therefore, the point is on the graph of To find another point on the line, any value can be substituted for in the formula. For simplicity, will be used. The graph of a direct variation passes through the
It has been found that the point
is also on the line. To draw the graph, these two points will be plotted and the line through them will be drawn.
This graph corresponds to Graph III. Note that in the context of the situation, a negative value of
does not make sense, since Tiffaniqua cannot run for a negative amount of hours.
minutes and how many hours it would take her to travel kilometers, the graph from Part B will be used. To find how many kilometers Tiffaniqua would travel in
It can be seen that, if Tiffaniqua jogged at a constant speed of
kilometers per hour, she would travel
minutes. Similarly, it would take her one and a half hours to travel
Since these two values are too large to be seen in the graph, the formula from Part A will be used.
is the distance in kilometers and
the time in hours. To find how many kilometers Tiffaniqua would travel in
minutes need to be expressed in hours.
hours. This value can be substituted for
in the equation.
minutes, Tiffaniqua would travel
kilometers. Finally, to calculate how many hours it would take her to travel
kilometers at this speed,
will be substituted for
in the formula.
If she kept her constant speed, it would take Tiffaniqua
hours to travel