Isa sa mahirap na parte ng pag-aaral sa Math ay ang pag-solve sa mga word problem dahil ito ay nangangailangan ng pag-aalisa o masusing paggamit ng common sense at imahinasyon. Ang mahirap na bahagi ng pagsagot sa mga word problem ay ang pag-setup ng equation. Para sa araling ito, ang ating pagtutuunan ng pansin ay ang pagsagot sa Distance Problem.
Ang "distance word problems" ay tinatawag din "uniform rate problems" kung saan ang isang bagay ay naglalakbay ng iisang bilis lamang o paiba-iba ang bilis sa isang paglalakbay. Kalimitan, ang isang distance problem ay sumasagot sa tanong na 1) kung gaano kabilis ang takbo; 2) gaano kalayo; o 3) gaano katagal.
Distance Formula
Ang distance formula ay d = rt,
kung saan ang d ay ang distansya; r ay ang uniform o average speed at ang t ay ang oras.
Babala sa paggamit ng distance formula.
Siguraduhin lamang na ang unit para sa oras (t) at distansya (d) ay PAREHO sa unit ng rate of speed (r).
Halimbawa kung ang rate (r) ay kilometer per hour, dapat ang distance ay kilometer at ang oras (time) ay hour.
Halimbawa ng Distance Problem:
My taxi is travelling at two different speeds from my house to the airport.On asphalt road, it travels at 105 mph. On concrete road, it travels at 115 mph. I left at 1:00 PM and arrived at the airport at 6:00 PM. The distance of my house from the airport is 555 miles. For how long did my taxi travel from each speed?
Solution:
Istratehiya sa pagsagot ng word problem:
1) Isaulo at ibigay ang general formula ng distance"
distance = rate of speed x time or d = rt
2) Dahil may dalawang speed ang taxi, ibigay ang bawat formula o equation nito.
a) Sa aspaltadong daan, ang speed o bilis (r) ng taxi ay 105 mph (miles per hour).
Dahil d = rt, ang equation nito ay
d = 105t
b) Sa sementadong daan, ang speed (r) ng taxi ay 115 mph.
Ang equation nito ay 555 - d = 115 ( 5-t).
Busisiin natin ang pangalawang equation.
Bakit ang distance (d) ay naging 555-d?
Ang total na distansya ng airport at bahay ay 555 miles (given). Dahil ang distance na itinakbo ng taxi sa aspaltadong daan ay d, ang matitirang distansyang tatakbuhin ng taxi sa sementadong daan ay 555 - d na lamang.
Saan kinuha ang 115(5-t)?
Ang 115 ay ang speed ng taxi sa sementadong daan. Bakit naging 5-t and oras ng taxi? Ang kabuuang oras ng paglalakbay ay 5 oras ( 6PM - 1PM = 5 hours). Dahil ang oras sa aspaltadong daan ay t na, natural lamang na ibawas ang t sa total na oras kaya naging 5 - t.
3) Pagsamahin o i-ADD ang dalawang equation.
Equation 1 : d = 105t
Equation 2: 555 - d = 115(5-t)
d + 555 - d = 105t + 115(5-t)
555 = 105t + 575 - 115t
555 = 575 - 10t
-20 = -10t
t = 2 hours
Ang t = 2 hours ay ang sagot sa unang equation na ang ibig sabihin ay 2 oras ang tinakbo ng taxi sa aspaltadong daan.
Ilang oras naman ang tinakbo ng taxi sa sementadong daan?
5 - t o 5 - 2 = 3 hours
Tatlong oras ang tinakbo ng taxi sa sementadong daan.
Hintayin ang ilang halimbawa para sa pagsagot ng distance problem.
Examples from Purple Math
Examples from Purple Math
- A boat travels for three hours with a current of 3 mph and then returns the same distance against the current in four hours. What is the boat's speed in calm water? How far did the boat travel one way?
| d | r | t | |
| downstream | d | b + 3 | 3 |
| upstream | d | b – 3 | 4 |
| total | 2d | --- | 7 |
(It may turn out that I won't need the "total" row.)
I have used "b" to indicate the boat's speed. Why are the rates "b + 3" and "b – 3"? Because I actually have two speeds combined into one on each trip. The boat has a certain speed (the "speed in calm water" that I'm looking for; this is the speed that registers on the speedometer), and the water has a certain speed (this is the "current"). When the boat is going with the current, the water's speed is added to the boat's speed. This makes sense, if you think about it: even if you cut the engine, the boat would still be moving, because the water would be carrying it downstream. When the boat is going against the current, the water's speed is subtracted from the boat's speed. This makes sense, too: if the water is going fast enough, the boat will still be going downstream (a "negative" speed, because the boat would be going backwards at this point), because the water is more powerful than the boat. (Think of a boat in a cartoon heading toward a waterfall. The guy paddles like crazy, but he still goes over the edge.)
| d | r | t | |
| downstream | d = 3(b + 3) | b + 3 | 3 |
| upstream | d = 4(b – 3) | b – 3 | 4 |
| total | 2d | --- | 7 |
Using "d = rt", the first row (of the revised table above) gives me:
- d = 3(b + 3)
The second row gives me:
- d = 4(b – 3)
Since these distances are the same, I will set them equal:
- 3(b + 3) = 4(b – 3)
In this case, I didn't need the "total" row.
- With the wind, an airplane travels 1120 miles in seven hours. Against the wind, it takes eight hours. Find the rate of the plane in still air and the velocity of the wind.
| d | r | t | |
| tailwind | 1120 | p + w | 7 |
| headwind | 1120 | p – w | 8 |
| total | 2240 | --- | 15 |
(I probably won't need the "total" row.) Just as with the last problem, I am really dealing with two rates together: the plane's speedometer reading, and the wind speed. When the plane turns around, the wind is no longer pushing the plane to go faster, but is instead pushing against the plane to slow it down. Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved
The first row gives me:
- 1120 = 7(p + w)
The second row gives me:
- 1120 = 8(p – w)
The temptation is to just set these equal, like I did with the last problem, but that just gives me:
- 7(p + w) = 8(p – w)
...which doesn't help much. I need to get rid of one of the variables.
I'll take that first equation:
- 1120 = 7(p + w)
...and divide through by 7:
- 160 = p + w
Then, subtracting w from either side, I get that p = 160 – w. I'll substitute "160 – w" for "p" in the second equation:
- 1120 = 8([160 – w] – w)
1120 = 8(160 – 2w)
- A spike is hammered into a train rail. You are standing at the other end of the rail. You hear the sound of the hammer strike both through the air and through the rail itself. These sounds arrive at your point six seconds apart. You know that sound travels through air at 1100 feet per second and through steel at 16,500 feet per second. How far away is that spike? (Round to one decimal place.)
| d | r | t | |
| air | d = 1100t | 1100 | t |
| steel | d = 16,500(t – 6) | 16,500 | t – 6 |
| total | --- | --- | 6 |
However long the sound took to travel through the air, it took six seconds less to propagate through the steel. (Since the speed through the steel is faster, then that travel-time must be shorter.) I multiply the rate by the time to get the values for the distance column. (Once again, I didn't need the "total" row.)
Since the distances are the same, I set the distance expressions equal to get:
- 1100t = 16,500(t – 6)
Solve for the time t, and then back-solve for the distance d by plugging t into either expression for the distance d.
Sana ay naunawaan.
