A question about average speed

In summary, the average speed is defined as the total distance covered divided by the amount of time elapsed. The formula for average speed is not the average of two speeds, but rather the total distance divided by the total time. This can be seen in an example where the "average of averages" formula does not give the correct answer.
  • #1
murshid_islam
458
19
If we travel s1 distance in t1 time and s2 distance in t2 time, is the average speed [tex]\frac{s_1 + s_2}{t_1 + t_2}[/tex] or [tex]\frac{\frac{s_1}{t_1} + \frac{s_2}{t_2}}{2}[/tex]
 
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  • #2
The first one: (total distance)/(total time).
 
  • #3
Doc Al said:
The first one: (total distance)/(total time).
Is there any reason why it isn't the second one or is it just defined that way?
 
  • #4
It's just what is meant by average speed.
 
  • #5
To follow up on what Doc Al said, average speed is defined as the total distance covered divided by the amount of time elapsed.

Here's an old problem that shows why the "average of averages" (as in your second formula) doesn't work.

On a certain road there is a hill that the road goes up and then down the other side. The uphill side is one mile and the downhill side is one mile. If I drive up the hill at an average speed of 30 mph, how fast do I need to drive down the other side to average 60 mph for the two miles?
 
  • #6
The "average of two numbers" is NOT, in general, the same as "average speed".
 
  • #7
I can see why you thought it might be the second one,

since [tex]\frac{s_1}{t_1}=v_1[/tex] and [tex]\frac{s_2}{t_2}=v_2[/tex] and the average of two speeds is [tex]\frac{v_1+v_2}{2}[/tex] then in conclusion the average speed over the whole distance would be your second one intuitively.

While it seems plausible, it's not correct.

There is a difference in these statements, for your problem, we want the average speed over the whole distance while for this problem, we just want the average of the two velocities. The two aren't the same because if we take [tex]s_2=1[/tex] but [tex]t_2[/tex] really small (approaching zero) such that [tex]v_2[/tex] is really large (approaching infinite), and [tex]s_1=t_1=v_1=1[/tex] then the average speed by the method above would be [tex](1+\infty)/2=\infty[/tex] but what it should really be is [tex](1+1)/(1+0)=2[/tex] since even though it travels really fast on the second journey, that journey is completed in such a short time that it doesn't mean the entire journey was completed in an infinitely small time (since we took a finite time to complete the first leg of the journey).
 
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  • #8
Mark44 said:
Here's an old problem that shows why the "average of averages" (as in your second formula) doesn't work.

On a certain road there is a hill that the road goes up and then down the other side. The uphill side is one mile and the downhill side is one mile. If I drive up the hill at an average speed of 30 mph, how fast do I need to drive down the other side to average 60 mph for the two miles?
:confused: I'm sorry, I still don't get it. Why does the second formula, i.e., [tex]\frac{\frac{s_1}{t_1} + \frac{s_2}{t_2}}{2}[/tex] or [tex]\frac{v_1 + v_2}{2}[/tex] not work here? If we use that formula, we get the velocity required downhill to be 90 mph. On the other hand, if I use the first formula, i.e., [tex]\frac{s_1 + s_2}{t_1 + t_2}[/tex], we have,

s1 = s2 = 1 mile
v1 = 30 mph
t1 = 1/30 h

Now, average speed = [tex]\frac{s_1 + s_2}{t_1 + t_2}[/tex]
[tex]60 = \frac{1+1}{\frac{1}{30} + t_2}[/tex]

[tex]t_2 = 0[/tex]

Then, [tex]v_2 = \frac{1}{0}[/tex] :confused:
 
  • #9
I believe Mark44 made an error in his problem statement. To average 60 mph, you'd have to complete the entire 2 mile trip in 2 minutes. But if you traveled the first half at 30 mph you've already used up 2 minutes. You'd have to travel the second mile in 0 time! (Infinite speed!) Which is impossible, of course.

But the formula works just fine--it tells you that you need to cover that second mile in 0 time. :eek:
 
  • #10
Doc Al said:
I believe Mark44 made an error in his problem statement. To average 60 mph, you'd have to complete the entire 2 mile trip in 2 minutes. But if you traveled the first half at 30 mph you've already used up 2 minutes. You'd have to travel the second mile in 0 time! (Infinite speed!) Which is impossible, of course.

But the formula works just fine--it tells you that you need to cover that second mile in 0 time. :eek:
But why exactly doesn't the second formula, i.e., [tex]\frac{v_1 + v_2}{2}[/tex] work here?

(I know that the books tell us that the formula for average speed is total distance divided by total time.I just want to know why it isn't the average of the two speeds, i.e., [tex]\frac{v_1 + v_2}{2}[/tex].)
 
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  • #11
murshid_islam said:
But why exactly doesn't the second formula, i.e., [tex]\frac{v_1 + v_2}{2}[/tex] work here?
Well, that formula gives the average of two speeds but it's not the average speed.

murshid_islam said:
:confused: I'm sorry, I still don't get it. Why does the second formula, i.e., [tex]\frac{\frac{s_1}{t_1} + \frac{s_2}{t_2}}{2}[/tex] or [tex]\frac{v_1 + v_2}{2}[/tex] not work here? If we use that formula, we get the velocity required downhill to be 90 mph.
But 90 mph isn't the right answer. If the second mile is traveled at a speed of 90 mph, the time it takes would be 1/90 hour. That would make the total time equal to:
1/30 + 1/90 = 4/90 hour.

But that makes the average speed equal to:
(2 miles)/(4/90 hour) = 45 mph (not 60, like it was supposed to be).
 

Related to A question about average speed

1. What is average speed?

Average speed is the total distance traveled divided by the total time taken to travel that distance. It is a measure of how fast an object is moving on average.

2. How is average speed calculated?

Average speed is calculated by dividing the total distance traveled by the total time taken to travel that distance. The formula for average speed is: average speed = total distance / total time.

3. Can average speed be negative?

Yes, average speed can be negative if an object is traveling in the opposite direction of its initial motion. For example, if a car travels north for 10 miles and then turns around and travels south for 10 miles, its average speed would be 0 miles per hour. However, if the car traveled north for 10 miles and then south for 20 miles, its average speed would be negative 10 miles per hour.

4. How is average speed different from instantaneous speed?

Average speed is the total distance traveled divided by the total time taken, while instantaneous speed is the speed at a specific moment in time. Average speed gives an overall picture of how fast an object is moving, while instantaneous speed can vary based on the object's acceleration or deceleration at any given moment.

5. What factors can affect average speed?

Average speed can be affected by various factors such as the distance traveled, the time taken to travel, the terrain, and any changes in speed or direction during the journey. Other factors such as traffic, weather conditions, and the type of transportation can also impact average speed.

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