I. Calculating an average speed
To calculate the average speed \(S\) of a vehicle when the distance \(d\) traveled and the duration \(t\) of the journey is known, we use the formula.
\[S = \frac{d}{t}\]
This formula can be algebraically manipulated to find the distance traveled if we know the duration and average speed of the journey (\(d = \frac{S}{t}\)) or the duration if we know the distance traveled and the average speed (\(t=\frac{d}{S}\)).
When carrying out this sort of calculation, be careful to always use matching units!
A. Example 1
Problem: A boat makes a journey of 28 km in 2 hours 20 minutes. What is its average speed?
Solution: Two important pieces of information are given: the distance and the duration. The journey time, however, is given in hours and minutes – either we commit to strictly hours or strictly minutes. Let's make the duration in hours only.
- Recall that 1 hours = 60 minutes, so:
\[\frac{1\;\mathrm{hour}}{60\;\mathrm{minutes}} = \frac{x}{20\;\mathrm{minutes}}\]
Solving this proportion problem, we get \(x = \frac{1}{3}\;\mathrm{hours}\).
All together, the duration in hours is \(2 + \frac{1}{3} = 2\;\frac{1}{3}\;\mathrm{hours}\).
Using the formula from \(S = \frac{d}{t}\), we can set \(d = 28\;\mathrm{km}\) and \(t = 2\;\frac{1}{3} \Rightarrow \frac{7}{3}\;\mathrm{hours}\). Notice that for the calculation, we have converted the mixed fraction to improper for simplicity (how?).
\[S = \frac{28\;\mathrm{km}}{\frac{7}{3}\;\mathrm{hours}}\]
\[S = \frac{12\;\mathrm{km}}{1\;\mathrm{hour}} \]
The journey is made at an average speed of 12 km/h.
B. Example 2
Problem: While cycling in the mountains, Martin travels 8 km up a hill, at an average speed of 8 km/h. He descends along the same route at an average speed of 32 km/h. What is his average speed for the round trip?
Solution: The round trip is 16 kilometers since Martin ascended and descended along the same route.
The total time when he ascended can be found using the formula \(t = \frac{d}{S}\).
\[t = \frac{8\;\mathrm{km}}{\frac{8\;\mathrm{km}}{1\;\mathrm{hour}}} \Rightarrow 1\;\mathrm{hour}\]
The total time when he descended can also be found using the formula \(t = \frac{d}{S}\).
\[t = \frac{8\;\mathrm{km}}{\frac{32\;\mathrm{km}}{1\;\mathrm{hour}}} \Rightarrow 0.25\;\mathrm{hour}\]
Thus, the total duration of the trip is 1 hr. + 0.25 hr. = 1.25 hr. Because the round trip is 16 kilometers, we can calculate the average speed using \(S = \frac{d}{t}\):
\[S = \frac{16\;\mathrm{km}}{1.25\;\mathrm{hr.}} \Rightarrow 12.8 \frac{\mathrm{km}}{\mathrm{hr.}}\]
The average speed of the round trip is 12.8 km/h.
Important Note: The average speed of the round trip is not the average of the two average speeds (that would be equal to \(\frac{32+8}{2}\), or 20 km/h).
II. Calculating a distance
When we know the distance and time, calculating the speed is a straightforward calculation since the variable \(S\) is already isolated on one side of the equation. Using some simple algebraic techniques, the formula can quickly be rearranged to be in terms of speed and time.
\[\require{cancel} S = \frac{d}{t}\]
Multiply both sides of the equation by \(t\):
\[(t)S = \frac{d}{t}(t)\]
This will cancel out the \(t\)'s on the right-side:
\[(t)S = \frac{d}{\cancel{t}}\cancel{(t)}\]
\[(t)S = d\]
\[\boxed{d = t \times S}\]
While you can memorize this formula and use it directly when needed, it is best that you simply memorize the speed formula and manipulate it whenever the variables (\(d\) or \(t\)) are required.
Problem: The speed of light is about 300,000 km/second. It takes about 8 minutes for light to get from the Sun to Earth. What is the distance of the planet Earth from the Sun?
Solution: The question tells information about the speed \((S)\) of light and the duration \((t)\) it is travelling.
\(S = 300,000 \frac{\mathrm{km}}{\mathrm{s}} \quad \mathrm{and} \quad t = 8\; \mathrm{min} \)
Because the units of speed do not match the duration unit, one must be made into the other. That is, either convert 300,000 km/second → km/minute or convert 8 minutes → seconds. We will convert the latter using the dimensional analysis technique introduced in a previous lesson:
- Recall that 1 minute = 60 seconds.
\[t = 8\;\cancel{\mathrm{min}} \times \frac{60\;\mathrm{s}}{1\;\cancel{\mathrm{min}}} = 480\;\mathrm{s}\]
Recall that the formula for speed is \(S = \frac{d}{t}\). Substituting into \(S\) and \(t\), our equation becomes:
\[\frac{300,000\;\mathrm{km}}{\mathrm{s}} = \dfrac{d}{480\;\mathrm{s}}\]
To solve for \(d\), we multiply both sides of the equation by \(\color{red}{480\;\mathrm{s}}\) to clear the denominator of the fraction on the right-side:
\[\color{red}{480\cancel{\mathrm{s}} \times}\;\frac{300,000\;\mathrm{km}}{\cancel{\mathrm{s}}} = \dfrac{d}{\cancel{480\;\mathrm{s}}} \color{red}{\times \cancel{480\;\mathrm{s}}}\]
\[480 \times 300,000\;\mathrm{km} = d\]
\[d = 144,000,000\;\mathrm{km}\]
Therefore, the distance of Earth from the Sun is about 144 million kilometers.
III. Calculating a time
Before we look at an example, let's manipulate the speed formula for \(t\). Just as did when we manipulated the equation for \(d\), we will attempt to clear the denominator by multiplying both sides by \(t\):
\[\require{cancel} S = \frac{d}{t}\]
Multiply both sides of the equation by \(t\):
\[(t)S = \frac{d}{t}(t)\]
This will cancel out the \(t\)'s on the right-side:
\[(t)S = \frac{d}{\cancel{t}}\cancel{(t)}\]
\[(t)S = d\]
Notice how \(t\) is currently being multiplied to \(S\). We want to "undo" the multiplication of \(S\) unto \(t\). Since the inverse operation of multiplication is division, we will divide both sides of the equation by \(S\):
\[\frac{(t)\cancel{S}}{\cancel{S}} = \frac{d}{S}\]
\[\boxed{t = \frac{d}{S}}\]
Problem: Natalie is going for a 15 km walk. She wants to get back at 4:30 pm and plans to walk at an average speed of 4 km/h. At what time should she set out?
Solution: First we can work out the journey time using the formula \(t = \frac{d}{S}\) (\(d\) is in km units and \(S\) is in km/h, so \(t\) will be in \(h\)); we replace \(S\) and \(d\) with their values, and get:
\[t = \frac{15\;\mathrm{km}}{4\frac{\mathrm{km}}{\mathrm{h}}}\]
The division statement will cause the km units to cancel out:
\[t = \frac{15\;\mathrm{\cancel{km}}}{4\frac{\mathrm{\cancel{km}}}{\mathrm{h}}}\]
\[t=3.75\;\mathrm{h}\]
Therefore, the walk will last three 3.75 hours.
Now we can calculate Natalie's departure time. For ease of subtraction, we will convert the clock time relative to 24 hours. So, 4:30 pm is equivalent to 16:30. Similarly, we will change the minutes (30) to hours (0.5), so 16.50 hours.
\[16.50\;\text{hours} - 3.75\;\text{hours} = 12.75\;\text{hours}\]
Notice that 12.75 is in hour units only. The whole number 12 implies that Natalie must have started at 12 PM. To convert the decimal portion (0.75 hours) to minutes, you can utilize the conversion factor of 1 hour = 60 minutes. Employing either the dimensional analysis technique previously taught in this course or relying on your own intuition, you can convert 0.75 hours to 45 minutes. Combining these two numbers gives the result 12 hours 45 minutes.
Therefore, Natalie must leave at 12:45 pm (or a quarter to one) to get back at 4:30 pm.
Tags
Speed Formula, Algebra, Average Speed, Formula Rerrangement
