Posts
It took me an hour to do the last two problems:
14. A motorist traveled from Town A to Town B. After traveling 1/3 of the distance for the journey at an average speed of 45 km/h, he continued to travel another 480 km to reach Town B. If his average speed for the entire journey was 54 km/h, what was his average speed for the last 2/3 of the distance? (answer: 60 km/h)
15. A car and a truck were traveling to Town Q at constant average speeds. The car overtook the truck when they were 420 km from Town Q. The car arrived at Town Q at 6:30 p.m. while the van was still 120 km away from Town Q. The van arrived at Town Q at 8:30 p.m. What was the average speed of the car? (answer: 84 km/h)
If I hadn't had the answers I would never have gotten the first problem right. I thought I should be able to set it up as a simple averageing problem:
(45 + 2r)/3 = 54
I did manage to work the problem correctly after an endless struggle. But I still don't (really) understand why a simple averaging of rates doesn't work, apart from a vague comprehension that I haven't "accounted for time."
The second one wasn't quite so taxing, but it was depressing nonetheless. I spent quite awhile trying to figure out exactly what I needed to compare to what -- I went down several wrong paths before suddenly realizing that I was missing the figure of 300 km.
Maybe I'll be smarter when I finally get some antibiotics.
I'm not holding my breath. (Coughing, yes. Holding my breath, no.)
If I had the energy I would now be panicking about whether I need to try to put C. back in the Singapore Math books.....actually, I would be alternating between panic and despair. He's only gotten through the first 20 lessons in Saxon Algebra 1/2; teaching a separate coherent curriculum around here is not happening, and is not going to happen.
It takes constant energy and vigilance to get him to do anything other than the work he's assigned at school, and his school assignments aren't teaching him what he needs to know.
AND -- I'm just about done wailing here -- winter illness and 7th grade wiliness have conspired to prevent C. from completing his fancy test prep book. The book has 49 lessons plus 6 "Progress Checks." The state test starts tomorrow and he's completed just 41 lessons and 2 Progress Checks.
He spent a good week or maybe two being too sick to work; then once he was better I was too sick to force him to work.
Now he's off at Stew Leonard's with his dad and I'm going to nap. When he gets back he's doing at least two more lessons; I've already told him he's probably going to have to read the others whether he does any of the problems or not.
I'm just not seeing how we get from here to calculus ---- or even from here to algebra 2.
department of corrections
This is the end of 5th grade test (pdf file).
The test linked to above is the test given at the end of 6A.
update: forty-two & david on distance-rate-time problems
from forty-two:
It helped me with problem 14 to remind myself that you don't need algebra to solve it. When I looked at it strictly arithmetically, it became a lot easier than it was at first look.
My thinking went like this:
Total distance = 480km / (2/3) = 720 km
Total time = 720km * 54km/hr = 13 1/3 hr
time for first 1/3 = (720km/3) * 45km/hr = 5 1/3 hr
time for last 2/3 = 13 1/3 hr - 5 1/3 hr = 8 hr
rate for last 2/3 = 480km / 8hr = 60 km/h
Honestly, I feel dumb, but I'm not sure WHY simple rate averaging doesn't work either. I know it doesn't, because I've beaten my head against enough rate problems to have first-hand experience with the many wrong ways to solve them, but I don't have a concrete reason WHY. I'll have to do some research.
david:
In order to take the average of two speeds, you need equal times, not equal distances.
For example, if I drive at 60 mph for one hour, then I drive at 50 mph for one hour, then my average speed is 55 mph. But if I drive at 60 mph for one mile, then I drive at 50 mph for one mile, then my average speed is less than 55 mph, because I spent more time at the slower speed.
For problem #14, you can use averages, provided that you express the rates as hours per kilometer. The equation is
(1/3)(1/45) + (2/3)(1/x) = 1/54
The idea is to put the unit of distance in the denominator, so we can take averages on the basis of equal distances instead of equal times. (I hope my explanation isn't too unclear.)
I'm going to read these explanations through a few times.
This is one of those cases where doing lots of these problems will inevitably lead to my experiencing these principles as obvious and natural.
procedural fluency preceding conceptual understanding
Happens a lot, I find.
_____________
state test coming right up (2006)
throwing money at the problem
more stuff only teachers can buy
help desk 1
state test coming right up (2007)
help desk 2
my life and welcome to it
inflammatory
canadianteacher.com
progress report
despair
28 out of 30
all the answers are belong to us
email to the math chair
second request
teacher's manual
it would be unusual
9 comments:
It helped me with problem 14 to remind myself that you don't need algebra to solve it. When I looked at it strictly arithmetically, it became a lot easier than it was at first look.
My thinking went like this:
Total distance = 480km / (2/3) = 720 km
Total time = 720km * 54km/hr = 13 1/3 hr
time for first 1/3 = (720km/3) * 45km/hr = 5 1/3 hr
time for last 2/3 = 13 1/3 hr - 5 1/3 hr = 8 hr
rate for last 2/3 = 480km / 8hr = 60 km/h
Honestly, I feel dumb, but I'm not sure WHY simple rate averaging doesn't work either. I know it doesn't, because I've beaten my head against enough rate problems to have first-hand experience with the many wrong ways to solve them, but I don't have a concrete reason WHY. I'll have to do some research.
In order to take the average of two speeds, you need equal times, not equal distances.
For example, if I drive at 60 mph for one hour, then I drive at 50 mph for one hour, then my average speed is 55 mph. But if I drive at 60 mph for one mile, then I drive at 50 mph for one mile, then my average speed is less than 55 mph, because I spent more time at the slower speed.
For problem #14, you can use averages, provided that you express the rates as hours per kilometer. The equation is
(1/3)(1/45) + (2/3)(1/x) = 1/54
The idea is to put the unit of distance in the denominator, so we can take averages on the basis of equal distances instead of equal times. (I hope my explanation isn't too unclear.)
I'm going to print these out and STUDY.
Thanks!
Ack! I made a mistake when I expanded my scratchwork into something more intelligible. On the second and third lines I wrote that I *multiplied* the distance by the rate to get time, when I really *divided*. I don't know what I was thinking - here are the revised calcs:
Total time = 720km / 54km/hr = 13 1/3 hr
time for first 1/3 = (720km/3) / 45km/hr = 5 1/3 hr
Sorry for any confusion!
This first problem is a great example of the power or bar modeling.
Since I can't draw it out, I'll have to explain it.
First, it helps to remember that a rate is distance/time. So to get the average rate for the second part of the trip, you'll need distance and time.
So if you draw the bar model rectangle and divide it into thirds, you'll see that 240km goes in each box. That means the first part of the trip was 240km, the second was 480 km and the whole is 720 km.
Now you need time for the second part of the trip. Since you have the rate for the whole trip and the rate for the first part of the trip, it is an easy calculation to the get the time for each part of the trip (see forty-two's response).
Finally, rate(secondpart)=distance(second part)/time(secondpart)
You know --- bar models aren't a "given" --- I drew a bar model and got the problem wrong.
(I'll draw yours now...)
You don't need algebra for #15 either.
Van arrived 2 hr later than the car (2030 vs 1830). But, when the car arrived at the endpoint, the van was still 120 km out. Therefore, the van's constant speed is 120 km / 2 hr = 60 km/hr.
The van and the car were at the same place and time 420 km from the end. For the van, that was 420 km / 60 km/hr = 7 hr. 7 hr earlier than 2030 (the van's arrival at the endpoint) is 1330. Since the car and van were at the same place and time, the car was also 420 km from the endpoint at 1330.
Therefore, the car traveled 420 km in (1830 - 1330) hr, so its constant speed is 420 km / 5 hr = 84 km/hr.
Whether you call them bar models or whatever -- I'd never heard of bar models until I hit KTM1 -- I always find it helpful to draw pictures and label what I know.
Both 14 and 15 are essentially drt problems. That is, they are testing your knowledge of (d)istance = (r)ate * (t)ime. You can add distances, or you can add times, but you cannot add rates.
If you forget whether it's d=rt or dr=t, remember that the units are your friends!
km = (km/hr) * (hr)
to enhance david's explanation:
d = rt
thus r = d/t (eqn 1)
Thus by the rules of adding fractions, we *have* to have equal times or stuff breaks.
Consider your problem.
Put t1 = time to travel first 1/3
Put t2 = time to travel next 2/3
Thus t = t1 + t2
Let r=54, r1, r2 and d, d1, d2 be defined similarly
We want to solve for r2 in our system where
r = 54 = d/t = (d1 + d2) / (t1 + t2) (eqn 2)
However, your method has you doing (in my notation)
(1/3) r1 + (2/3)r2 and solving for r2. However, that is (substituting)
= (1/3) (d1/t1) + (2/3)(d2/t2)
= (1/3) [ dt/t1 + 2d2/t2 ]
= (d1t2 + 2d2t1) / (3 t1 t2)
which is wrong on inspection
HTH
earl
Post a Comment