Carcass wrote:
Blanca took a non-stop car trip that encompassed three different sections of roadways. Each of the three sections covered the same distance. Blanca averaged 45 miles per hour over the first section, 60 miles per hour over the second section, and 54 miles per hour for the entire trip. What was her average speed for the third section to the nearest mile per hour?
First, we can jot down what we know.
D_Total = Total Distance
R_Average = Average Speed of entire trip
T_Total = Total time of entire trip
The total trip can be described with:
D_Total = (R_Average)(T_Time)Each segment of the trip has the same distance, so we'll say the distance of each segment can de denoted as
D.
So we can also say that
D_Total = 3D.
We don't know how long they were travelling on each segment, so we'll say:
t_1 = time spent travelling on first segment
t_2 = time spent travelling on second segment
t_3 = time spent travelling on third segment
Finally, we want to find the speed on the third segment, we'll call that
x.
So according to all the information we have the following equations:
3D = (54)(T_Time)D = (45)(t_1)D = (60)(t_2)D = (x)(t_3)At this point, we could try some algebra that could get complicated. Instead, we have two unknowns (the distance and the time) in each equation. Since the distance is the same for each segment, we can choose any arbitrary number for the distance, and we can work our way to
x from there. I picked
360 as the distance for each sector because it's a multple of both 45 and 60, making the math a bit easier.
So from here:
(3)(360) = (54)(T_Time)360 = (45)(t_1)360 = (60)(t_2)360 = (x)(t_3)Simplifying:
1080 = (54)(T_Time)20 = (T_Time)8 = (t_1) 6 = (t_2)360 = (x)(t_3)Note that the total time can be expressed as: t_1 + t_2 + t_3
So we have:
20 = 8 + 6 + t_320 = 14 + t_36 = t_3Now we can finally solve for x:
360 = (x)(t_3)360 = (x)(6)60 = (x)Therefore, the speed of the third segment must be 60mph.