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September 23, 2011 IOT POLY ENGINEERING I1-22 DRILL:

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Presentation on theme: "September 23, 2011 IOT POLY ENGINEERING I1-22 DRILL:"— Presentation transcript:

1 September 23, 2011 IOT POLY ENGINEERING I1-22 DRILL: DESCRIBE HOW THE AXES OF A GRAPH ARE LABELED PROPERLY. DESCRIBE HOW DATA POINTS ON A GRAPH ARE PROTECTED FROM OBLITERATION WHEN DRAWING A LINE OR CURVE.

2 IOT POLY ENGINEERING I1-22 HERE IS THE COMPLETED CHART AND GRAPH FROM YESTERDAY’S PRINTED HANDOUT SHEET.

3 Will the trains collide? Explain.
IOT POLY ENGINEERING I1-22 The engineer of an express train traveling at 30 m/s notices a freight train 84 meters up ahead traveling in the same direction at only 10 m/s. The engineer immediately puts on the brakes and the train decelerates at 2 m/s/s. Will the trains collide? Explain. If there is a collision, how many seconds after the engineer puts on the brakes will the trains collide? If there is not a collision, how far apart will the trains be when the express train finally stops?

4 The distance traveled by the freight train is given by: D = 84 + 10 t
IOT POLY ENGINEERING I1-22 This problem is very difficult, but can be solved with algebra and a graph. We need to determine if both trains are ever in the same place at the same time. We will calculate the distance of each train based upon the location of the express train when the brakes were applied. The distance traveled by the freight train is given by: D = t The distance traveled by the express train is given by: D = 30 t – t2

5 IOT POLY ENGINEERING I1-22 The express train loses 2 m/s of speed every second, so we could assume that it will lose all 30 m/s of its original speed within 15 seconds. Now, let’s calculate where each train is located at t=15 seconds. D = t D = (15) = = 234 meters D = 30 t – t2 D = 30 (15) – (152) = 450 – 225 = 225 meters The problem stated: “If there is not a collision, how far apart will the trains be when the express train finally stops?” Based on our assumption that the express train will stop in 15 seconds, we conclude that there is not a collision and the trains are 9 meters apart. Now, let’s solve the same problem using a graph.

6 IOT POLY ENGINEERING I1-22 HERE IS THE COMPLETED CHART AND GRAPH FROM YESTERDAY’S PRINTED HANDOUT SHEET. For every second of time, determine the location of each train from the original location of the express train. The freight train starts at 84 meters, while the express train begins at 0 meters. Based on the graph of each train’s location, what is your conclusion? The trains collide at t=6 s.

7 TEMPERATURE CONVERSION DERIVATION
IOT POLY ENGINEERING I1-22 TEMPERATURE CONVERSION DERIVATION This is the normal boiling temperature of water at sea level. 212 100 F C This line represents any temperature on the thermometer. F - 32 C - 0 32 We can set up a proportion: This is the normal freezing temperature of water at sea level. F - 32 C - 0 = We can now solve for either F or C. Let’s do both!

8 = = = I1-22 F - 32 C - 0 212 - 32 100 - 0 C F - 32 Cross Multiply 180
IOT POLY ENGINEERING I1-22 F - 32 C - 0 = C F - 32 = Cross Multiply 180 100 Divide by GCF (greatest common factor) = 100(F – 32) 180C 5(F – 32) = 9C 5(F – 32) = 9C 5(F – 32) = 9C F – 32 = 9C 5(F – 32) = C 5 9 This equation is used to solve for C when F is known. F = 9C + 32 5 This equation is used to solve for F when C is known.

9 HOMEWORK - September 23, 2011 I1-22 373 100 K C 273
IOT POLY ENGINEERING I1-22 HOMEWORK - September 23, 2011 373 100 K C 273 A. BASED ON THIS DIAGRAM, DERIVE 2 EQUATIONS: TO CONVERT KELVIN TO CELSIUS TO CONVERT CELSIUS TO KELVIN. B. CAN THE FAHRENHEIT TEMPERATURE EVER MATCH THE CELSIUS TEMPERATURE? ( X F = X C ) USE ALGEBRA TO PROVE YOUR CONCLUSION.


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