Speed, Distance, and Time

A common set of physics problems ask students to determine either the speed, distance, or travel time of something given the other two variables. These problems are interesting since they describe very basic situations that occur regularly for many people. For example, a problem might say: "Find the distance a car has traveled in fifteen minutes if it travels at a constant speed of \(75 \text {km/hr}\)." Often in these problems, we work with an average velocity or speed, which simplifies the laws of motion used to calculate the desired quantity. Let's see how that works.

As long as the speed is constant or average, the relationship between speed, distance, and time is expressed in this equation

\[\mbox{Speed} = \frac{\mbox{Distance}}{\mbox{Time}},\]

which can also be rearranged as

\[\mbox{Time} = \frac{\mbox{Distance}}{\mbox{Speed}}\]

and

\[\mbox{Distance} = \mbox{Speed} \times \mbox{Time}.\]

Speed, distance, and time problems ask to solve for one of the three variables given certain information. In these problems, objects are moving at either constant speeds or average speeds.

Most problems will give values for two variables and ask for the third.

Bernie boards a train at 1:00 PM and gets off at 5:00 PM. During this trip, the train traveled 360 kilometers. What was the train's average speed in kilometers per hour?

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In this problem, the total time is 4 hours and the total distance is \(360\text{ km},\) which we can plug into the equation: \[\mbox{Speed} = \frac{\mbox{Distance}}{\mbox{Time}}= \frac{360~\mbox{km}}{4~\mbox{h}} = 90~\mbox{km/h}. \ _\square \]

When working with these problems, always pay attention to the units for speed, distance, and time. Converting units may be necessary to obtaining a correct answer.

A horse is trotting along at a constant speed of 8 miles per hour. How many miles will it travel in 45 minutes?

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The equation for calculating distance is \[\mbox{Distance} = \mbox{Speed} \times \mbox{Time},\] but we won't arrive at the correct answer if we just multiply 8 and 45 together, as the answer would be in units of \(\mbox{miles} \times \mbox{minute} / \mbox{hour}\). To fix this, we incorporate a unit conversion: \[\mbox{Distance} = \frac{8~\mbox{miles}}{~\mbox{hour}} \times 45~\mbox{minutes} \times \frac{1~\mbox{hour}}{60~\mbox{minutes}} = 6~\mbox{miles}. \ _\square \]
Alternatively, we can convert the speed to units of miles per minute and calculate for distance: \[\mbox{Distance} = \frac{2}{15}~\frac{\mbox{miles}}{\mbox{minute}} \times 45~\mbox{minutes} = 6~\mbox{miles},\]
or we can convert time to units of hours before calculating: \[\mbox{Distance} = 8~\frac{\mbox{miles}}{\mbox{hour}} \times \frac{3}{4}~\mbox{hours} = 6~\mbox{miles}.\]
Any of these methods will give the correct units and answer. \(_\square\)

In more involved problems, it is convenient to use variables such as \(v\), \(d\), and \(t\) for speed, distance, and velocity, respectively.

Albert and Danny are running in a long-distance race. Albert runs at 6 miles per hour while Danny runs at 5 miles per hour. You may assume they run at a constant speed throughout the race. When Danny reaches the 25 mile mark, Albert is exactly 40 minutes away from finishing. What is the race's distance in miles?

\[\] Let's begin by calculating how long it takes for Danny to run 25 miles: \[\mbox{Time} = \frac{\mbox{Distance}}{\mbox{Speed}}= \frac{25~\mbox{miles}}{5~\mbox{miles/hour}}= 5~\mbox{hours}.\] So, it will take Albert \(5~\mbox{hours} + 40~\mbox{minutes}\), or \(\frac{17}{3}~\mbox{hours}\), to finish the race. Now we can calculate the race's distance: \[\begin{align}
\mbox{Distance} &= \mbox{Speed} \times \mbox{Time} \\
&= (6~\mbox{miles/hour}) \times \left(\frac{17}{3}~\mbox{hours}\right) \\
&= 34~\mbox{miles}.\ _\square
\end{align}\]

A cheetah spots a gazelle \(300\text{ m}\) away and sprints towards it at \(100\text{ km/h}.\) At the same time, the gazelle runs away from the cheetah at \(80\text{ km/h}.\) How many seconds does it take for the cheetah to catch the gazelle?

\[\] Let's set up equations representing the distance the cheetah travels and the distance the gazelle travels. If we set distance \(d\) equal to \(0\) as the cheetah's starting point, we have \[\begin{align}
d_\text{cheetah} &= 100t \\
d_\text{gazelle} &= 0.3 + 80t.
\end{align}\] Note that time \(t\) here is in units of hours, and \(300\text{ m}\) was converted to \(0.3\text{ km}.\)

The cheetah catches the gazelle when \[\begin{align}
d_\text{cheetah} &=d_\text{gazelle} \\
100t &= 0.3 + 80t \\
20t &= 0.3 \\
t &= 0.015~\mbox{hours}.
\end{align}\] Converting that answer to seconds, we find that the cheetah catches the gazelle in \(54~\mbox{seconds}\). \(_\square\)