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(Video) Distance, Rate, and Time Word Problem

by PatrickJMT

This video by PatrickJMT works through a word problem involving distance, rate, and time.

$\large{\text{Summary}}$

Distance, rate, and time are connected to one another by the equation: $$\text{distance}=\text{rate}\times\text{time}\\ d=rt$$ For example, if your rate is $60\text{ mph}$ and you drive for $3\text{ hours}$, you will have driven $60\times3=180\text{ miles}$.

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Example

The speed of a boat in still water is $10\text{ mph}$. It travels $24\text{ miles}$ upstream and $24\text{ miles}$ downstream in $5\text{ hours}$. What is the speed of the current?

These can be tricky to solve but take it one step at a time and set up equations. First, let's define the speed of the current as $c$. $$c=\text{speed of current}$$ Next, we'll find the distance, rate, and time for when the boat goes upstream and downstream.

Upstream

When the boat travels upstream, it's fighting against the current, so its speed (or rate) is reduced: $$r_{\text{up}}=10-c$$ The distance it travels upstream is $24\text{ miles}$. $$d_{\text{up}}=24$$ We know both trips combined take $5\text{ hours}$ but we don't know how long the upstream trip takes. Let's just call it $t_{\text{up}}$ for now.

Downstream

Next, when going downstream, the boat works with the current so its speed is $10+c$. $$r_{\text{down}}=10+c$$ Again, it travels $24\text{ miles}$. $$d_{\text{down}}=24$$ And we don't know the time, so we'll call it $t_{\text{down}}$.

Time

We aren't given the time for each trip, but we are given the total trip's time. It takes $5\text{ hours}$ to travel upstream and downstream. We can write this as: $$t_{\text{up}}+t_{\text{down}}=5$$ Using the distance equation, $d=rt$, we can solve for $t$ and get: $$t=\frac dr$$ Plugging this into our time equation, we get: $$\frac{d_{\text{up}}}{r_{\text{up}}}+\frac{d_{\text{down}}}{r_{\text{down}}}=5$$

Current

Finally, we plug in the information we have and solve for $c$ (find the rate). $$\begin{align} \frac{d_{\text{up}}}{r_{\text{up}}}+\frac{d_{\text{down}}}{r_{\text{down}}}&=5\\ \frac{24}{10-c}+\frac{24}{10+c}&=5\\ {\color{#8925AE}{(10-c)(10+c)}}\left(\frac{24}{10-c}+\frac{24}{10+c}\right)&={\color{#8925AE}{(10-c)(10+c)}}(5)\\ 24(10+c)+24(10-c)&=5(100-c^2)\\ 240+24c+240-24c&=500-5c^2\\ 480&=500-5c^2\\ 5c^2&=20\\ c^2&=4\\ c&=2 \end{align}$$ The current is $2\text{ mph}$.

(Video) Linear Distance, Rate, and Time Word Problem

by PatrickJMT

In this video, Patricks works through a word problem involving distance, rate, and time.

$\large{\text{Summary}}$

Distance, rate, and time are connected to one another by the literal equation: $$\text{distance}=\text{rate}\times\text{time}\\ d=rt$$ For example, if your rate is $60\text{ mph}$ and you drive for $3\text{ hours}$, you will have driven $60\times3=180\text{ miles}$.

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Example

Marjorie was getting blackberries along a path from her back door to the rear of her garden. On the way to the rear of the garden, she walked at an average speed of $8\text{ feet per minute}$, and on the way back to her back door, she walked at an average speed of $16\text{ feet per minute}$. If the round trip took $21\text{ minutes}$, how far in feet is it from her back door to the rear of her garden.

In this problem we want to find the distance between her door and her garden. To do this, we want to look at her trip to the garden, and her trip back.

To the Garden

Going to the garden, she walks $8\text{ feet per minute}$. $$r_{\text{garden}}=8$$ We don't know the time it took to walk to the garden, only the time it took to walk there and back. So, we'll use $t_{\text{garden}}$.

To the House

Going back to the house, she walks $16\text{ feet per minute}$. $$r_{\text{house}}=16$$ Again, we don't know the time, so we'll call it $t_{\text{house}}$.

Time

We're given the time for the entire trip, $21\text{ minutes}$. Wecan write this equation as: $$t_{\text{garden}}+t_{\text{house}}=21$$

Using the distance equation, $d=rt$, we can solve the equation for $t$. $$t=\frac dr$$

Plugging this into our time equation, we get: $$\frac{d}{r_{\text{garden}}}+\frac{d}{r_{\text{house}}}=21$$

Solve for Distance

Finally, to find the distance, we plug in our known values and solve this equation. $$\begin{align} \frac{d}{8}+\frac{d}{16}&=21\\ \frac{2d}{16}+\frac{d}{16}&=21\\ \frac{3d}{16}&=21\\ 3d&=336\\ d&=112 \end{align}$$ The distance between her house and garden is $112\text{ feet}$.

Rate-time-distance is pretty easy to understand, the formula to use for rate-time-distance is: $$\text{Distance} = \text{Rate} \times \text{Time}$$ $$\text{Rate} = \frac{\text{Distance}}{\text{Time}}$$ $$\text{Time} = \frac{\text{Distance}}{\text{Rate}}$$

Let's try a problem that uses the third equation, finding the time.

Two cars are traveling from the same point, at $5$ A.M., traveling in opposite directions at $40$ and $50\text{ mph}$ respectively. At what time will they be $450\text{ miles}$ apart? Assume neither car changes time zones.

After $t$ hours the distances $D_1$ and $D_2,$ in miles per hour, traveled by the two cars are given by $$\begin{align} D_1 &= 40t\\ D_2 &= 50t \end{align}$$

After $t$ hours, the distance $D$ separating the two cars is given by $$\begin{align} D &= D_1 + D_2\\ &= 40t + 50t\\ &= 90t \end{align}$$

Distance $D$ will be equal to $450$ miles when: $$D = 90 t = 450 \text{ miles}$$

To find the time $t$ for $D$ to be $450$ miles, solve the above equation for $t$ to obtain $$t = 5\text{ hours}$$

After five hours pass, it will be $10$ A.M., as it was $5$ A.M. when we started.

Let's try another problem, this time involving a train instead of cars.

A train drove $2888$ miles starting at its station, always going at $80$ mph. What is the amount of time since it left the station?

We know, that the train in $2888$ miles away from its station, we also know that the rate it is going at is $80$ miles per hour. We use the formula:

$$\text{Time} = \frac{\text{Distance}}{\text{Rate}}.$$ $$\frac{2888}{80}= 36.1$$

hours since the train left from its station.

We could also use the formula, $\text{Distance} = \text{rate} \times \text{time}.$ Let's try a problem using this one!

The biker is going to a parade at a speed of $25$ mph , she has biked for $2$ hours. What is the distance she traveled from her house to the parade?

The rate is $25$ mph and the time is $2$ hours so our equation becomes

$$D = 25 \ \text{mph} \ \times 2 \ \text{hours} \ = \boxed{50} \ \text{miles}.$$

How are Rate, Time, and Distance Related?

These $3$ variables—rate, time, and distance—are all closely related to each other. They're like the Three Musketeers of math. If you know the values of two of them, you can easily find the third.

Below, we'll take a closer look at how to work with them.

The expression below equates the $3$ variables. We'll manipulate this to solve a few example problems together.

When working with this equation, make sure that the units of all three variables are consistent.

For example, if your distance is in meters and your time is in seconds, your rate should have meters per second as its units.

Let's work on some practice problems.

Look at the chart below to see how we got the answer.

Let's plug in the values we have to the general equation:

The problem asks us directly about distance. So, let's start by plugging in the values we already have to the general equation. Then, we can solve for distance.

Now, we can solve for distance.

If we want to find the relationship between distance, rate, and time, we use a simple formula.

Work and rate problems are common applications of rational expressions.

We can use this formula to solve problems involving distance, rate, and time. To use the formula, you will plug in the values you know and solve for the values you don't know.

You may be wondering what happens when both coordinates depend on time. When both coordinates depend on time, the motion can be modeled using parametric equations. Don't worry, you'll learn about this concept later!

We know that: $$d=3$$ $$t=2$$ So we can plug these values into the distance-rate-time equation. $$d=rt$$ $$3=r(2)$$ $$\frac32=r$$