Solving problems with the formula for distance, rate, and time

Formula that relates distance, rate, and time

This lesson looks at how to find distance, rate and time given two out of three of these values.

Distance, rate and time are related by the equation

$\text{Distance}=\text{Rate} \cdot \text{Time}$

$D=RT$

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Let’s talk about the units of each of these values.

Distance has units in inches, feet, miles, or centimeters, meters, kilometers etc.

Time has units in seconds, minutes, hours, etc.

Rate has units in distance/time, for example inches/second, miles/hour, or kilometers/hours.

Before you can use the formula $D=RT$ you need to make sure that your units for the distance and time are the same units in your rate. If they aren’t, you’ll need to change them so you’re working with the same units.

How to solve distance, rate, and time problems

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Finding average rate given distance and time

Example

Heather ran $56$ km in $5$ hours. What was Heather’s average rate in km/hr?

We’ll use the formula for distance.

$\text{Distance}=\text{Rate} \cdot \text{Time}$

$D=RT$

Let’s write down what we know.

$D=56$ km

$T=5$ hr

If we plug these into the distance formula, we get

$D=RT$

$56\text{ km} = R\cdot 5\text{ hr}$

Now solve for the Rate.

$\frac{56\ \text{km}}{5\ \text{hr}} = \frac{R \cdot 5\ \text{hr}}{5\ \text{hr}}$

$R=11.2\ \frac{\text{km}}{\text{hr}}$

Before you can use the formula D=RT you need to make sure that your units for the distance and time are the same units in your rate.

Distance, rate, and time problems with two people who leave at different times

Example

Susan and Benjamin were $60$ miles apart on a straight trail. Susan started walking toward Benjamin at a rate of $5$ mph at 7:30 a.m. Benjamin left three hours later and they met on the trail at 3:30 p.m. How fast is Benjamin?

We’ve been given information about distance, rate and time, so we’ll use the formula

$\text{Distance}=\text{Rate} \cdot \text{Time}$

$D=RT$

where $D$ is the distance traveled, $R$ is the rate, and $T$ is the time. We can use subscripts to create unique equations for Susan and Benjamin.

Susan: $D_{S} =R_{S} T_{S}$

Benjamin: $D_{B} =R_{B} T_{B}$

We know that in order to meet each other, they must have covered a distance of $60$ miles between them. Therefore,

$D_{S}+D_{B}=60$

Since we know that $D_{S}=R_{S}T_{S}$ and $D_{B}=R_{B}T_{B}$, we can make substitutions into this equation for their rates and times, instead of their distances.

$R_{S}T_{S}+R_{B}T_{B}=60$

The problem tells us that Susan was walking at $5$ mph, and that she walked for $8$ hours, since she walked from 7:30 a.m. until 3:30 p.m. So

$(5)(8)+R_{B}T_{B}=60$

$40+R_{B}T_{B}=60$

$R_{B}T_{B}=20$

Benjamin left three hours after Susan, which means he started walking at 10:30 p.m., and he kept walking until they met at 3:30 p.m., which means he walked for $5$ hours. So

$R_{B}(5)=20$

$R_{B}=4$

Which means that Benjamin walks at $4$ mph.

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