![Page 1: Math as Problem-Solving w Active Physics asks students to problem-solve w Some of this problem-solving involves math w Some of the math is encoded in the](https://reader036.vdocument.in/reader036/viewer/2022082820/56649ea95503460f94bacac6/html5/thumbnails/1.jpg)
Math as Problem-SolvingMath as Problem-Solving
Active Physics asks students to problem-solve
Some of this problem-solving involves math
Some of the math is encoded in the language of word problems.
![Page 2: Math as Problem-Solving w Active Physics asks students to problem-solve w Some of this problem-solving involves math w Some of the math is encoded in the](https://reader036.vdocument.in/reader036/viewer/2022082820/56649ea95503460f94bacac6/html5/thumbnails/2.jpg)
What do you do when you don’t know what to do?
rereading the problem using what you already know looking for patterns applying the solution to a similar problem making a picture of diagram
![Page 3: Math as Problem-Solving w Active Physics asks students to problem-solve w Some of this problem-solving involves math w Some of the math is encoded in the](https://reader036.vdocument.in/reader036/viewer/2022082820/56649ea95503460f94bacac6/html5/thumbnails/3.jpg)
Word Problem Context
Home Module, Student Text, page 56. Power = current x voltage P = IV
![Page 4: Math as Problem-Solving w Active Physics asks students to problem-solve w Some of this problem-solving involves math w Some of the math is encoded in the](https://reader036.vdocument.in/reader036/viewer/2022082820/56649ea95503460f94bacac6/html5/thumbnails/4.jpg)
Issues for the Uninitiated
Need for units• “I’ll take 100?”• one hundred of what?
Units and variables• P power in Watts (W)• I current in Amperes (A)• V voltage in Volts (V)
![Page 5: Math as Problem-Solving w Active Physics asks students to problem-solve w Some of this problem-solving involves math w Some of the math is encoded in the](https://reader036.vdocument.in/reader036/viewer/2022082820/56649ea95503460f94bacac6/html5/thumbnails/5.jpg)
The Word Problem
How much current, in amps, must be flowing through the filament of a 60-W light bulb when it is operating in a 120-V household circuit? A 100-W light bulb? Show your calculations in your log.
What are some issues that this problem raises for the uninitiated?
![Page 6: Math as Problem-Solving w Active Physics asks students to problem-solve w Some of this problem-solving involves math w Some of the math is encoded in the](https://reader036.vdocument.in/reader036/viewer/2022082820/56649ea95503460f94bacac6/html5/thumbnails/6.jpg)
What’s the problem?
What’s unknown? What is given? What mathematical relationship do I use?
![Page 7: Math as Problem-Solving w Active Physics asks students to problem-solve w Some of this problem-solving involves math w Some of the math is encoded in the](https://reader036.vdocument.in/reader036/viewer/2022082820/56649ea95503460f94bacac6/html5/thumbnails/7.jpg)
Solution: Substitution
P = I V 60W = A? 120V What number when multiplied by 120 gives
60?
![Page 8: Math as Problem-Solving w Active Physics asks students to problem-solve w Some of this problem-solving involves math w Some of the math is encoded in the](https://reader036.vdocument.in/reader036/viewer/2022082820/56649ea95503460f94bacac6/html5/thumbnails/8.jpg)
Solution: Rearrangement
P = I V Divide both sides by V
P / V = I V / V The result is
I = P / V Now substituting
? = 60 W / 120 V
VIVVP // VIVVP //