Speed, Distance, and Time

Definition

The relationship between speed, distance, and time is expressed in this equation

Speed=DistanceTime,\mbox{Speed} = \frac{\mbox{Distance}}{\mbox{Time}},

which can also be rearranged as

Time=DistanceSpeed\mbox{Time} = \frac{\mbox{Distance}}{\mbox{Speed}}

and

Distance=Speed×Time.\mbox{Distance} = \mbox{Speed} \times \mbox{Time}.

Technique

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

Most problems will give values for two variables and ask us to solve for the third. For example:

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?

In this problem, the total time is 4 hours and the total distance is 360 km, which we can plug into the equation: Speed=DistanceTime=360 km4 h=90 km/h.□\begin{aligned} \mbox{Speed} &= \frac{\mbox{Distance}}{\mbox{Time}}\\ &= \frac{360~\mbox{km}}{4~\mbox{h}}\\ &= 90~\mbox{km/h}. \quad_\square \end{aligned}

When working with these problems, always pay attention to the units for speed, distance, and time. Here is a problem where we need to convert units to get the 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?

The equation for calculating distance is Distance=Speed×Time,\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 miles×minute/hour\mbox{miles} \times \mbox{minute} / \mbox{hour}. To fix this, we incorporate a unit conversion. Distance=8 miles hour×45 minutes×1 hour60 minutes=6 miles□\begin{aligned} \mbox{Distance} &= \frac{8~\mbox{miles}}{~\mbox{hour}} \times 45~\mbox{minutes} \times \frac{1~\mbox{hour}}{60~\mbox{minutes}} \\ &= 6~\mbox{miles} \quad_\square \end{aligned}
Alternatively, we can convert the speed to units of miles per minute and calculate for distance: Distance=215 milesminute×45 minutes=6 miles\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: Distance=8 mileshour×34 hours=6 miles.\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.

In more involved problems, we may find it convenient to use variables such as vv, dd, and tt for speed, distance, and velocity respectively.

Application and Extensions

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. Let Time=DistanceSpeed=25 miles5 miles/hour=5 hours.\begin{aligned} \mbox{Time} &= \frac{\mbox{Distance}}{\mbox{Speed}}\\ &= \frac{25~\mbox{miles}}{5~\mbox{miles/hour}}\\ &= 5~\mbox{hours}. \end{aligned} So, it will take Albert 5 hours+40 minutes5~\mbox{hours} + 40~\mbox{minutes}, or 173 hours\frac{17}{3}~\mbox{hours}, to finish the race. Now we can calculate the race's distance: Distance=Speed×Time=(6 miles/hour)×(173 hours)=34 miles□\begin{aligned} \mbox{Distance} &= \mbox{Speed} \times \mbox{Time} \\ &= (6~\mbox{miles/hour}) \times (\frac{17}{3}~\mbox{hours}) \\ &= 34~\mbox{miles}\quad_\square \end{aligned}

 

A cheetah spots a gazelle 300 m away and sprints towards it at 100 km/h. At the same time, the gazelle runs away from the cheetah at 80 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 dd equal to 00 as the cheetah's starting point, we have: dcheetah=100tdgazelle=0.3+80t.\begin{aligned} d_\text{cheetah} &= 100t \\ d_\text{gazelle} &= 0.3 + 80t. \end{aligned} Note that time tt here is in units of hours, and 300 m was coverted to 0.3 km.

The cheetah catches the gazelle when dcheetah=dgazelle100t=0.3+80t20t=0.3t=0.015 hours.\begin{aligned} d_\text{cheetah} &=d_\text{gazelle} \\ 100t &= 0.3 + 80t \\ 20t &= 0.3 \\ t &= 0.015~\mbox{hours}. \end{aligned} Converting that answer to seconds, we find that the cheetah catches the gazelle in 54 seconds54~\mbox{seconds}.□_\square

#Algebra #Distance #Speed #Time #KeyTechniques

Note by Arron Kau
7 years, 2 months ago

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