On Friday I posted this puzzle.

You drive to your office at 20 mph and return by the same route at 30 mph. Discounting the time spent at the office, what was your average speed?

If you have not tried to solve it, have a go now. For everyone else the answer is after the break.

The average speed is 24 mph, but can you explain why?

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called **PUZZLED** and is available for the **Kindle **(UK here and USA here) and on the **iBookstore** (UK here in the USA here). You can try 101 of the puzzles for free here.

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Because you drove effectively 0 miles in a lot of time, your average speed is 0 miles per hour 😉

If you calculate it as Richard did, assume you drive 60 miles to your office. 3 hours to your office, 2 hours back, that’s 5 hours for 120 miles. That’s an average of 24 mph.

I think you mean average velocity, which measures the rate of change of position, rather than average speed.

For a more general solution that applies to all distances: It’s all about the relationship between 20 and 30 mph. You spend 60% of the time traveling at 20 mph, but you only spend 40% of the time traveling at 30 mph — because you’re home faster.

So 20 mph contributes with 60% and 20 mph with 40% to the total speed: 20 mph *60% + 30 mph *40% = 24 mph

Extra stuff, poorly explained: 25 mph, the average of 20 mph and 30 mph, is wrong because you didn’t spend half and half of your time traveling at those speeds. But you might say “but I spent half and half of the distance traveling at those speeds!”, and well, that’s because it’s miles PER HOUR: So the time spent is crucial. Miles per hour is just one way to represent speed. Another way is “hours per mile”. 20 mph = 0.05 hpm, and 30 mph = 0.33 hmp, and the average of them is 0.042 hmp, which is 24 mph! So averaging “hours per mile works”, when the distances are equal, and averaging “miles per hour” works if the times are equal.

That is clever, Kristian.

The question explicitly discounts the time not travelling.

Yes, got that. In short: 1/(0.5/20+0.5/30) = 24

Longer answer:

Half the distance was done with 20 mph, so took 0.5/20 = 0.025 “time units”

The other half was done with 30 mph, so took 0.5/30 = 0.0166666666666667 “time units”.

Together the total distance costed 0.041666666666666667 “time units”

So the average speed is 1/0.04166666666666667 = 24 mph

Can I? Yes. Will I? No.

I have nothing to say on this one. It’s all been said.

Me too

Unfortunately, Eddie and Jono, a lot of the other contributors to this thread don’t feel the same way as you do.

I feel the same way, so I will also say nothing.

I made it 25mph; is that in the margin of error?

No, that’s applying a wrong calculation.

First, i also came to the 25 but it was to easely…

The point is that the definition is Miles per Hours what means “Distance” per “Time”. But the time of the both ways is not identical. That should lets you to be suspicious and to start to examine it.

We should be able to explain why it is not 25 that is v average = (V1 + V2)/2.

One way is to say that if you come back at the speed of the light, the average speed is not half the speed of the light, in fact if you make no time to come back it gives you an average speed of 50 mph. So you see that the formula is not (V1+V2)/2.

It should exist an easy way to find the average speed without calculus

Suppose the route is 120 miles long.

go at 20mph it takes 6 hours

return at 30mph it takes 4 hours

The total distance done is 240 miles in 10 hours : average speed 240/10 = 24

My drive to the office is slower (say 20mpg as in Richard’s example) because I have to spend time finding somewhere to park and then walk. My drive home is faster (say 30mpg) because the space outside my house is free.

This means my actual driving time is the same in both directions (25mpg) even though the journey time is different (average 24mpg as Richard notes).

This is a good paradox for discussing speed verses velocities.

Strictly speaking the average velocity in this puzzlw is zero.

Speed describes only how fast an object is moving, whereas velocity gives both how fast and in what direction the object is moving. If a car is said to travel at 20 mph, its speed has been specified. However, if the car is said to move at 20 mph to the north, its velocity has now been specified. The big difference can be noticed when we consider movement around a circle. When something moves in a circle and returns to its starting point its average velocity is zero but its average speed is found by dividing the circumference of the circle by the time taken to move around the circle. This is because the average velocity is calculated by only considering the displacement between the starting and the end points while the average speed considers only the total distance traveled.

The big difference can be noticed when we consider movement around a circle. When something moves in a circle and returns to its starting point its average velocity is zero but its average speed is found by dividing the circumference of the circle by the time taken to move around the circle. This is because the average velocity is calculated by only considering the displacement between the starting and the end points while the average speed considers only the total distance traveled.

Your comments about speed vs. velocity are correct, and would be relevant if this puzzle had asked for average velocity.

A=Average in mph.

X is the distance driven in miles to work. Both numbers are greater than zero.

The total time in hours spent driving is 2X Miles/ A mph , that is the total distance driven divided by the average distance.

The total time spent driving must also be equal to the time spent driving to work plus the time spent driving back from work, which comes to X miles / 20 mph+X miles/ 30 mph

2X/A=X/20+X/30 =>

2/A=1/20+1/30 =>

2/A = 50/600 =>

2/A = 1/12 =>

A/12 = 2>

A=24

If you realize to points, the calculation is easy

First – the distance traveled cancels out, and second, the sum 1/20+1/30, sums up to the inverse of *half* the average, so the one calculation needed to do the job is

2/(1/20+1/30)=24

Thank you -this was a clear answer.

where can ye view your book 101 friday puzzles online with answers?

please any one send me the link 🙂

Go to the Friday posting of this puzzle. All the links are there

(2*20*30)/(20+30)=600*2/50=24

To Anonymous, yes i went to the friday posting of this puzzle but i could not find the link of the answers of all the ‘101 friday puzzles’ book which you can view online.

Why are people posting the same calculations over and over again. Roland’s post is sufficient albeit incorrect. The average speed is (20+30)/2 = 25mph.

End of. Let’s move on with our lives

You are wrong. The definition of average speed is not averaging two different speeds, it’s the total distance traveled divided by the total time elapsed whilst traveling. If you are trying to shut down a debate, at least get your definitions correct. Here’s a link to a fairly elementary explanation.

http://www.bbc.co.uk/schools/gcsebitesize/science/add_ocr_gateway/forces/speedrev1.shtml

Note that speed is a scalar, not a vector, so those arguing that the average speed is zero are wrong too.

nb. in doing this calculation we are implicitly working it out as relative to the frame of reference of the Earth’s surface. The Earth is, of course, moving in a complicated

Imagine the distance is 60 miles. at 20 miles per hour it takes 3 hours, at 30 miles her hour it takes 2 hours. so overall it takes 5 hours to go 120 miles which is reduced to 24 miles per hour. It cannot be an average of the two speeds because mph is a ratio not a constant.

What is definition of speed? Speed is a distance over the time. Based on definition average speed will be total distance over total time, where total time will be time commuting to office (t1) plus time commuting from office (t2). Total distance equals sum of distance to office (s1) and distance from office (s2). Since s1=s2 (name it just s). t1 is not equal to t2. Again I will apply speed definition to t1 and t2.

v1=s/t1; v2=s/t2

Or t1=s/v1; t2=s/v2

Then average speed is v=(s1+s2)/(t1+t2)

Combine all equesions to one:

v=2*s/(s/v1 + s/v2) it will be redused by s.

v= 2/(1/v1 + 1/v2) or v= 24 mph.

distance = rate * time

and 20 t1 = 30 t2 since distance into work = distance home

After division by 30,

(2/3) t1 = t2 . Now we know the time home is 2/3 of the time into work.

average velocity = total distance / total time

average velocity = (20 t1 + 30 t2) / (t1 + t2)

After substitution,

average velocity = (20 t1 + 20 t1) / (t1 + (2/3) t1) = 40 t1/ ((5/3) t1) = 40/(5/3) = 24

Thank you.

Can one take the same route from office to home as one took from home to the office?

Unless one works from home…….av. speed = 0

@Lazy T

You are making the mistake of treating speed as a vector physical property. It’s not, speed is a scalar. You might have a case if you are talking about average velocity, but that’s not what the question asked.

http://van.physics.illinois.edu/qa/listing.php?id=19188

thanks SJ,

so if I just sit here then what’s my average speed?

I think the issue here is ambiguous English : the phrase “Unless …” could do with “…in which case,,,” in the middle to clarify what was meant. (Otherwise it could be read as “Apart from the case when one works from home…”)

soz for the lazy comment, the pedantic point was

if you go to work, (a to b to c to d), you can’t come home by the same route, you’d go d to a. If however you work from home you go ( a to a) and return (a to a) ie so you then could go at 20mph for 0 sec and return at 30mph for 0 sec giving an average speed of 0mph.

I’ve lots of time on my hands this am, so I’m going to cover a balloon in pepper.

Considering the time taken to travel 1 mile at each speed:

20mph = 3 minutes; 30mph = 2 minutes.

Total time per 2 miles (1 mile each way) on the return trip = 5 minutes. Multiply by 12 to get an hour and we have (12×2=24) miles per (12×5=60) minutes, or 24mph.

Those who claim it is 25mph are incorrect. To get an average of 25mph. the journey times would have to be constant and distances variable, but the question clearly gives constant distances (return trip) and variable times.

In general,

Distance = Rate x Time (or d= rt); therefore,

(eq. 1) t = d/r, and

(eq. 2) r = d/t

So…

Time to go to work = d/20 ….(eq, 1)

Time to go home = d/30 ….(eq. 1)

Time to go both ways = d/20 + d/30 = 3d/60 + 2d/60 =5d/60 = d/12

Distance both ways = 2d

Rate (speed) = 2d / (d/12) ….(eq. 2)

…which simplifies to just 24 (d cancels out).

What’s dr/dt then?

Everything about the right answer that needs to be said has been said. I will add a good rule to use in problems like this. Many did apply this rule, perhaps without realizing it. As stated, it is clear that the distance from home to office is irrelevant to the answer. So you can just make up any distance you want. Easiest distance is some multiple of 20 and 30, and the least common multiple is 60. Just assume that the distance from home to office is 60. That makes the solution very simple: total distance traveling to and from the office is 120 miles with a total time of 5 hours (3 hours to office, 2 hours return) and there’s your 24 mph.

24mph is the harmonic mean of 20 & 30

That’s not a coincidence. Harmonic means are used in physics to average rates.

Think of it this way: If you go 1 MPH on the first trip and it’s a mile, it takes 1 hour. Right?

Now you go 60 MPH on the trip back (a mile), and that’s 1 minute.

So now you’ve gone 2 miles in 1 hour, 1 minute. That’s an average speed of 2 MPH.

This is an amazing car! It accelerates and decelerates in 0 seconds. That has got be hell on your tires.

I got about 38.6, or 10.7

What? You don’t actually expect me to do maths in non-SI units???

What’s the answer in kilometers per hour?

Another random thought. What do the average speed cameras calculate?

I liked this puzzler because it’s good for reminding people about how their instincts can lead them astray. Most people would answer 25 mph without thinking it through.

I’m guessing they calculate average speed

That was meant as a reply to the speed camera question. They time you between two points. Distance between the two points divided by time taken gives the average speed.

We calculate average velocity as – 2V1*V2/V1+V2 So putting values we get – 2*20*30/20+30 =24mph