# What is the distance formula problem for achieving an average speed of 50 mph?

• hackedagainanda
In summary, the problem states that Mr. Mercer wants to drive his Raceabout two miles at an average speed of 50 miles per hour. He drives the first mile at 25 miles per hour, leaving him with a certain amount of time to drive the second mile. To achieve an average speed of 50 miles per hour, he needs to drive the second mile at a rate of 1.5 minutes, which would equal 75 miles per hour. Therefore, the answers to the given questions are: 1. 1.5 minutes and 2. Yes, at 75 miles per hour.
hackedagainanda
Homework Statement
Mr. Mercer wants to drive his Raceabout two miles at an average speed of 50 miles per hour.
1.If he drives the first mile at 25 miles per hour , how much time does he have left to drive the second mile?

2. Can he do it, If so, at what speed?
Relevant Equations
Distance = D, rate of speed = r, and time = t, D=rt
I'm pretty lost on this one. I think he needs to drive the second mile at 75 miles per hour to average 50 miles per hour. So he needs to drive the second mile at a rate of 1.5 minutes which would = 75/1.5 = 50

So my answer for 1. 1.5 minutes, and my answer for 2. is Yes, at 75 miles per hour. I'm fairly confident about my answer to 2. but feel very uncertain about the answer to 1.

hackedagainanda said:
... I'm pretty lost on this one. I think he needs to drive the second mile at 75 miles per hour to average 50 miles per hour...
What makes you assume that?
Hint: "... he drives the first mile at 25 miles per hour".

I thought it was 75 + 25 = 100, Then divide that by 2 to get an average of 50.

That would work if half of the time were run at 25 mph and the second half of the time were run at 50 mph, being that time the one used if moving at constant 50 mph.
The problem tells you that 25 mph was the speed during half of the total distance of two miles.

Is it like 2= 2(25+x)/(25+x)? Where x = the speed of the second half of the distance?

Sorry, I can't make sense of that relation.
We have a rate, named speed, which is distance/time.

The original condition of the problem is:
50 mph = 2 miles / x hour

Then, they give you the second condition:
25 mph = 1 mile / y hour

Then, they ask you about the time (z hour) that would take the car to complete the second mile at a different rate of speed (the one you have calculated as 75 mph).

If you calculate x time first, you will have the range of available time that Mr. Mercer has to reach his goal of average speed.

I'm still extremely lost. I get (1/25) for x and y. And 1/25 of an hour is 2.5 minutes. So is it 25 mph for the second mile?

OK.
You have calculated that 0.04 hour or 2.4 minutes is the time that Mr. Mercer would need to use in order to drive his Raceabout two miles at an average speed of 50 miles per hour.

How much time he uses driving the first mile of that 2-mile trip at 25 miles per hour?
How much time he would use driving the second mile of that 2-mile trip at estimated 75 miles per hour?
What the average speed of that trip would be, being that average speed equal to (distance 1/time 1) + (distance 2/time 2)?

Hint:
If you normally drive the 30 miles that separate your home from your school at an average speed of 30 mph, it takes you one hour.
One day, traffic is worse than normal and you can only drive at steady 15 mph.
One hour after departing, which is the normal time the trip takes, how many miles have you covered?

The average speed is given by ##50=\frac{d}{t}##. Another way to look at this is by decomposing both ##d## and ##t## into an addition of two quantities representing each driving phase.
$$50=\frac{d_1+d_2}{t_1+t_2}$$
You have ##d_1## and ##d_2##, and you can find ##t_1##.

I have crossed that because I was getting zero and I don't think that that is correct, but if we want to have an average speed of 50 miles per hour, then that is the same as saying that we want ##\frac{50}{1}=\frac{2}{2/50}=\frac{2}{1/25}## miles per hour, or ##2## miles per ##1/25## hour. Though he already used that ##1/25## of an hour in his first mile.

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Lnewqban and Delta2
hackedagainanda said:
Homework Statement:: Mr. Mercer wants to drive his Raceabout two miles at an average speed of 50 miles per hour.
1.If he drives the first mile at 25 miles per hour , how much time does he have left to drive the second mile?

2. Can he do it, If so, at what speed?
Relevant Equations:: Distance = D, rate of speed = r, and time = t, D=rt

I'm pretty lost on this one. I think he needs to drive the second mile at 75 miles per hour to average 50 miles per hour. So he needs to drive the second mile at a rate of 1.5 minutes which would = 75/1.5 = 50

So my answer for 1. 1.5 minutes, and my answer for 2. is Yes, at 75 miles per hour. I'm fairly confident about my answer to 2. but feel very uncertain about the answer to 1.
EDIT: You want the speed that will give you the correct value of 50 mph for total_distance/total_time. You already know that total_distance=2 mi. So the first step is to determine total_time.

You should approach this in a simple, step-by-step process. How much time will it take to drive 2 miles at 50 mph? That tells you how much time you have to work with. Then how much time have you used in driving 1 mile at 25 mph? How much time is left for driving the remaining mile?

As @Lnewqban points out, "average" is tricky because there are different ways of averaging (time average or distance average or just arithmetic average of two numbers). Your answer is a certain type of speed average (arithmetic), but it will not go the distances and times that this problem specifies.

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Lnewqban and SammyS
Mr. Mercer wants to drive his Raceabout two miles at an average speed of 50 miles per hour.
1.If he drives the first mile at 25 miles per hour , how much time does he have left to drive the second mile?

2. Can he do it, If so, at what speed?
You understand this example better if you arrange the information into tabular form.

Code:
                                     Rate               Time                Distance

First mile                        25                                          1

Second mile                       v                                            1

Combination                                                                2
Now you fill in the other information using your basic relationship of D=rt, or t=D/r.
I have chosen v instead of "r" for the unknown speed for the second mile.

(How do we make this stuff line-up better?)

FactChecker
@FactChecker arithmetic average as in ##(a_1+a_2)/2=a_{avg}##?
That would make the op's answer correct, then. $$50=\frac12(\frac{1}{1/25}+\frac1t)\Leftrightarrow t=1/75\,\mathrm h$$

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archaic said:
The average speed is given by ##50=\frac{d}{t}##. Another way to look at this is by decomposing both ##d## and ##t## into an addition of two quantities representing each driving phase.
$$50=\frac{d_1+d_2}{t_1+t_2}$$
You have ##d_1## and ##d_2##, and you can find ##t_1##.

I have crossed that because I was getting zero and I don't think that that is correct, but if we want to have an average speed of 50 miles per hour, then that is the same as saying that we want ##\frac{50}{1}=\frac{2}{2/50}=\frac{2}{1/25}## miles per hour, or ##2## miles per ##1/25## hour. Though he already used that ##1/25## of an hour in his first mile.
You shouldn't cross this in my opinion it was correct. It is just that it turns out that ##t_2=0## and ##d_2=1## which means that he has to run the second mile with infinite speed ##\frac{d_2}{t_2}=\infty##. But the problem asks if it is possible, so i guess the answer here is NO it isn't possible (unless we argue that infinite speed is possible).

I took the average speed to mean average over time. If it is average over distance then 75 is the correct answer. And same answer of 75 if it means the arithmetic average of two numbers.

archaic said:
@FactChecker arithmetic average as in ##(a_1+a_2)/2=a_{avg}##?
That would make the op's answer correct, then. $$50=\frac12(\frac{1}{1/25}+\frac1t)\Leftrightarrow t=1/75\,\mathrm h$$
The arithmetic average does not apply in this problem. If Mr. Mercer drives the first mile at 25 mi/hr and the second mile at 75 mi/hr, his average rate for the two miles is 37.5 mi/hr, not 50 mi/hr.

archaic, Lnewqban and FactChecker
archaic said:
@FactChecker arithmetic average as in ##(a_1+a_2)/2=a_{avg}##?
That would make the op's answer correct, then. $$50=\frac12(\frac{1}{1/25}+\frac1t)\Leftrightarrow t=1/75\,\mathrm h$$
Yes, and that is my point. Although that is called an "average", it does not go the correct distance in the correct time. The desired average is the speed that gives the desired value for total_distance/total_time. Since we know the value of total_distance, the first step is to determine total_time.

archaic and Lnewqban
If you look at post #11 again, the sum for the missing time information, using v as the unknown speed of second mile, 1/25+1/v.

If you want the average speed for the whole TWO miles be 50, then 2/(1/25+1/v)=50.
Solve this equation for v.

I am finding NO SOLUTION.

Lnewqban said:
OK.
You have calculated that 0.04 hour or 2.4 minutes is the time that Mr. Mercer would need to use in order to drive his Raceabout two miles at an average speed of 50 miles per hour.

How much time he uses driving the first mile of that 2-mile trip at 25 miles per hour?
How much time he would use driving the second mile of that 2-mile trip at estimated 75 miles per hour?
What the average speed of that trip would be, being that average speed equal to (distance 1/time 1) + (distance 2/time 2)?

Hint:
If you normally drive the 30 miles that separate your home from your school at an average speed of 30 mph, it takes you one hour.
One day, traffic is worse than normal and you can only drive at steady 15 mph.
One hour after departing, which is the normal time the trip takes, how many miles have you covered?
From this hint I am getting the the trip takes 2 hours for the 30 miles. Applying this to my problem if the total time needs to = 1/25 an hour and at a speed of 25 mph for the first mile = 1/25 the answer for 1 is no time left to complete the second mile. Then the answer for 2 is no it isn't possible, no matter how fast you go.

Lnewqban and FactChecker
hackedagainanda said:
From this hint I am getting the the trip takes 2 hours for the 30 miles. Applying this to my problem if the total time needs to = 1/25 an hour and at a speed of 25 mph for the first mile = 1/25 the answer for 1 is no time left to complete the second mile. Then the answer for 2 is no it isn't possible, no matter how fast you go.

hackedagainanda
Thanks everyone so much! I miss having a teacher to ask these questions to...

Another way you might look at the problem is if you travel 2 miles at 50 mph, how far would you travel (in the same time) at 25 mph ?

The way presented in post #11 is a typical way that a beginning algebra student would commonly be taught to analyze this kind of constant travel rates exercise and to solve.

FactChecker
symbolipoint said:
(How do we make this stuff line-up better?)
When prepping your text for inclusion in the [CODE] [/CODE] section, use a text editor which does not use a proportional font. I used notepad. Then trust the preview pane and ignore the goofiness of the editing pane.
Code:
                                     Rate               Time                Distance

First mile                           25                                       1

Second mile                          v                                        1

Combination                                                                   2

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jbriggs444 said:
use a text editor which does not use a proportional font. I used notepad. Then trust the preview pane and ignore the goofiness of the editing pane.
Maybe I will soon try that. No information comes in Notepad telling which fonts are proportional or not proportional.

symbolipoint said:
Maybe I will soon try that. No information comes in Notepad telling which fonts are proportional or not proportional.
I always go with Courier if I want a fixed font.

Checking just now under Info => Help => BB codes I see that one can use Courier here on PF:

[FONT=Courier New]This is a fixed font[/FONT]
[FONT=Courier New]iiiiiiiiiiiiiiiiiiii[/FONT]
[FONT=Courier New]WWWWWWWWWWWWWWWWWWWW[/FONT]

and display in a fixed font without using a [CODE] section.

In order to get the "[FONT]" to display as text above I used the trick of putting the "[" inside a color tag. If you click Reply and hit the gear icon on the right end of the toolbar (Toggle BB code) you can see inside that technique. The easy way to do the color tag is with the paint drop icon from the tool bar.

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## 1. What is the distance formula problem?

The distance formula problem is a mathematical equation used to calculate the distance between two points in a coordinate plane. It is often used in physics and engineering to determine the distance an object travels over a given time period.

## 2. How is the distance formula problem related to achieving an average speed of 50 mph?

The distance formula problem is essential for calculating the distance an object travels to achieve an average speed of 50 mph. By plugging in the values for distance and time into the formula, the average speed can be determined.

## 3. Can the distance formula problem be used for any type of motion?

Yes, the distance formula problem can be used for any type of motion, whether it is linear, circular, or even a combination of both. As long as the motion can be represented on a coordinate plane, the distance formula can be applied.

## 4. What are the variables used in the distance formula problem?

The variables used in the distance formula problem are distance (d), time (t), and speed (s). Distance is measured in units such as meters or miles, time is measured in units such as seconds or hours, and speed is measured in units such as meters per second or miles per hour.

## 5. How can the distance formula problem be applied in real-life situations?

The distance formula problem can be applied in various real-life situations, such as calculating the distance a car travels on a road trip, determining the distance a plane travels during a flight, or even finding the distance between two cities on a map. It is a useful tool in many fields, including transportation, navigation, and sports.

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