# How to solve uniform motion problems

## The distance formula for uniform motion problems

In this lesson we'll look at how to compare and solve for different values in the ???\text{Distance}=\text{Rate} \cdot \text{Time}??? equation when you have related scenarios.

Uniform motion explains the distance of an object when it travels at a constant speed, the rate, over a period of time.

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To compare different rates, times, and distances you can use subscripts to keep track of which pieces go with which equation.

Remember the formula:

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

???D=RT???

## How to solve uniform motion problems step-by-step

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## Two trains leave the same station

**Example**

A train leaves Station ???A??? at a constant speed and arrives at Station ???B??? in ???8??? hours. A second train leaves Station ???A??? at a constant rate of ???40??? mph and arrives at Station ???B??? in ???10??? hours. What was the constant speed of the first train?

Since both trains are traveling from Station ???A??? to Station ???B???, they’re traveling the same distance. We can recognize that this is a uniform motion problem.

We’ll use the formula,

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

???D=RT???

where ???D??? is the distance each of them traveled, ???R??? is the constant speed of the trains, and ???T??? is the time it took them to from station to station. We can use subscripts to create unique equations for each train. We’ll call them Train 1 and Train 2.

Train 1: ???D_1=R_1T_1???

Train 2: ???D_2=R_2T_2???

Let’s organize the information we know about each train.

Train 1:

???D_1=????

???R_1=????

???T_1=8??? hours

Train 2:

???D_2=????

???R_2=40??? mph

???T_2=10??? hours

We also know that ???D_1=D_2??? because the trains traveled the same distance.

Now let’s plug this information into the ???D=RT??? equations.

???D_1=R_1T_1???

???D_1=R_1(8\ \text{hrs})???

and

???D_2=R_2T_2???

???D_2=(40\ \text{mph})(10\ \text{hrs})???

???D_2=400\ \text{miles}???

Remember both trains traveled the same distance from Station ???A??? to Station ???B???. Since their distances are equal, we can say that ???D_1=D_2???, and plug the value we just found for ???D_2??? into the ???D=RT??? equation for Train 1.

???400\ \text{miles}=r_1(8\ \text{hrs})???

???50\ \text{mph} = r_1???

The first train traveled at a constant speed of ???50\ \text{mph}??? from Station ???A??? to Station ???B???.

Uniform motion explains the distance of an object when it travels at a constant speed, the rate, over a period of time.

Let’s do one more example.

## Two drivers depart from the same location

**Example**

Cassie is driving at a constant rate of ???30??? mph on the highway. ???4??? hours later, her friend Susan starts from the same point and drives at a constant rate of ???60??? mph and passes Cassie. How many hours in the car does it take each girl to be at the same spot on the highway? And how far did each girl travel?

When they’re at the same spot on the highway, they’ll have both traveled the same distance. We can recognize this is a uniform motion problem. We can use the formula ???D=RT??? where ???D??? is the distance each of them traveled, ???R??? is the rate at which they traveled, and ???T??? is the time it took them to get there. We can use subscripts to set up a model for each woman’s travel.

Cassie: ???D_c=R_cT_c???

Susan: ???D_s=R_sT_s???

The problem tells us that Cassie traveled at a rate of ???30??? mph, that Susan traveled at a rate of ???60??? mph, and that it took Susan ???4??? hours less than Cassie to travel the same distance (because she left ???4??? hours later).

Let’s set up what we know.

Cassie:

???D_c=????

???R_c = 30\ \text{mph}???

???T_c=????

???D_c=(30\ \text{mph})T_c???

Susan:

???D_s=????

???R_s = 60\ \text{mph}???

???T_s=T_c-4???

???D_s=(60\ \text{mph})(T_c-4)???

Remember, Cassie and Susan both traveled the same distance. Since their distances are equal, we can say that ???D_c=D_s??? and get

???(30\ \text{mph})T_c=(60\ \text{mph})(T_c-4)???

???\frac{(30\ \text{mph})T_c}{60\ \text{mph}}=\frac{(60\ \text{mph})(T_c-4)}{60\ \text{mph}}???

???\frac{1}{2}T_c=T_c-4???

???\frac{1}{2}T_c-T_c=T_c-T_c-4???

???-\frac{1}{2}T_c=-4???

???-2\cdot-\frac{1}{2}T_c=-2\cdot-4???

???T_c=8??? hours

We can now say that it took Cassie ???8??? hours to travel the distance. So it took Susan ???4??? hours less than it took Cassie to go the same distance.

???T_s=T_c-4???

???T_s=8-4???

???T_s=4??? hours

Now that we have a rate and a time for both Cassie and Susan, we can verify that they traveled the same distance.

Cassie:

???D_c=R_cT_c???

???D_c=30(8)???

???D_c=240??? miles

Susan:

???D_s=R_sT_s???

???D_s=60(4)???

???D_s=240??? miles