# Average Speed

• Mar 15th 2009, 05:40 PM
magentarita
Average Speed
Lenore drove 200km. For the first 100km of the trip, she drove at a speed of 40 km/h. For the second half of the trip, she traveled at a speed of 60 km/h. What was the average speed of the entire trip?

MY WORK:

I know we can use R = D/t

I first need to find the time.

T = D/R

T = 100/40

T = 2.5 hours

I then did the same thing for the second part of the trip.

T = 100/60

T = 1.7 hours

This is where I get stuck.

• Mar 15th 2009, 06:04 PM
e^(i*pi)
Quote:

Originally Posted by magentarita
Lenore drove 200km. For the first 100km of the trip, she drove at a speed of 40 km/h. For the second half of the trip, she traveled at a speed of 60 km/h. What was the average speed of the entire trip?

MY WORK:
I know we can use R = D/t
I first need to find the time.
T = D/R
T = 100/40
T = 2.5 hours
I then did the same thing for the second part of the trip.T = 100/60T = 1.7 hoursThis is where I get stuck.[/COLOR]

[/SIZE]

Using all those colour tags makes it hard to read the problem

The answer is simply the mean of the two speeds since she travelled an equal distance at each speed:

$\displaystyle \frac{40+60}{2} = 50km/h$

The full answer is below:

$\displaystyle \frac{100km \times 40km/h + 100km \times 60km/h}{200km} = \frac{100km}{200km} (40+60)km/h = 50km/h$

The algebraic answer would be
$\displaystyle s_n$ = distance covered on the nth leg
$\displaystyle v_n$ = velocity on the nth leg

$\displaystyle \frac{s_1v_1 + s_2v_2 + s_3v_3 + ... + s_nv_n}{s_1 + s_2 + s_3 + s_n}$
• Mar 15th 2009, 06:17 PM
magentarita
ok...
Quote:

Originally Posted by e^(i*pi)
Using all those colour tags makes it hard to read the problem

The answer is simply the mean of the two speeds since she travelled an equal distance at each speed:

$\displaystyle \frac{40+60}{2} = 50km/h$

The full answer is below:

$\displaystyle \frac{100km \times 40km/h + 100km \times 60km/h}{200km} = \frac{100km}{200km} (40+60)km/h = 50km/h$

The algebraic answer would be
$\displaystyle s_n$ = distance covered on the nth leg
$\displaystyle v_n$ = velocity on the nth leg

$\displaystyle \frac{s_1v_1 + s_2v_2 + s_3v_3 + ... + s_nv_n}{s_1 + s_2 + s_3 + s_n}$

I made it harder than it really is.
• Mar 16th 2009, 09:49 AM
stapel
Quote:

Originally Posted by magentarita
Lenore drove 200km. For the first 100km of the trip, she drove at a speed of 40 km/h. For the second half of the trip, she traveled at a speed of 60 km/h. What was the average speed of the entire trip?

While in some cases the rates can be averaged, this is not true in general! Instead, play it safe, and use the standard set-up for "uniform rate" word problems:

[HTML]+--------+-----+----+--------+
|########| d = r * t |
+--------+-----+----+--------+
| first | 100 | 40 | |
+--------+-----+----+--------+
| second | 100 | 60 | |
+--------+-----+----+--------+[/HTML]
As you mention, we can use "d = rt" in the form of "t = d/r" to find the times:

[HTML]+--------+-----+----+--------+
|########| d = r * t |
+--------+-----+----+--------+
| first | 100 | 40 | 100/40 |
+--------+-----+----+--------+
| second | 100 | 60 | 100/60 |
+--------+-----+----+--------+[/HTML]
Simplifying, we get times of 2 1/2 hours and 1 2/3 hours, for a total time of 4 1/6 hours (or 4 hours and 10 minutes).

Note: By using fractions, I have avoided the round-off error involved in "converting" 5/3 to "1.7". (Wink)

The average rate for the entire trip is of course the total miles divided by the total time: (200 mi)/(25/6 hr) = (200 * 6)/(25) mph = 8 * 6 mph = 48 mph. This makes sense, since a longer time will have been spent at the slower speed, making the overall average closer to the lower rate of travel.

Had the times been equal, rather than the distances, averaging might have had a better chance of giving the correct answer. :D