alittlebitofcalculus-pdf-january2011-111112001007-phpapp01

Ds infinitesimal piece of the cur ve yx ds dx x1 0

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Unformatted text preview: 30 25 m/sec 25 S P E E D E D 20 15 10 5 A (0 , 0 ) 5 B 10 15 TIME (sec) C 20 25 t DISTANCE = SPEED * TIME [meters = meters * seconds] seconds over the interval [ t = 10 secs, t = 20 secs] the SPEED = 25 m/sec ( constant ). Distance travelled from t = 10 secs to t = 20 secs is 25m/sec ∗ 10 sec = 250m which is the ar ea under the speed cur v e = area of rectangle BCDE. area curv ar In the case of distance travelled between t = 0 secs and t = 10 secs, the travelled SPEED is NOT constant. What SPEED are you going to use? 181 By looking at the graph we know the SPEED at each and every INSTANT between t = 0 secs and t = 10 secs. So we have to take the product of SPEED and TIME at each and ever y INSTANT and sum it up. This in effect will give the ar ea under the speed cur v e from t = 0 secs to t = 10 secs area ar which is the area of triangle ABE. This is 1 base * height = 1 * 10 secs * 25 m / s ec = 125 meters . --2 2 Conceptually we note that from the speed cur v e ( INSTANTANEOUS RATE speed OF CHANGE of position graph) we got the CHANGE in position. CHANGE in position position has both sign and magnitude. In this case we have denoted going East (right) as positiv neg tiv positive and going West (left) as negative on the x-axis. We may denote the SPEED ne of the car by x’(t) and hence its POSITION on the x-axis will be x(t). Because the acceleration is steady or constant over [0, 10) we may use the average acceleration over [0, 10). x’(10) − x’(0) 2.5m /sec2 acceleration over [0, 10) : x”(t) = = 10 secs acceleration over t ≥ 10 : x”(t) = 0 2.5t2/ over [0, 10] speed over [0, 10] : 2 x(t) = 2.5t m/ 25t − 125 for t ≥ 10 x’(t) = ∫x”(t)dt = sec speed over t ≥ 10 : 375 x’(t) = 25 m/sec 250 position over [0, 10] : 2 125 x(t) = ∫ 2.5t dt = 2.5t / 2 position over t ≥ 10 : [meters] { x(t) = ∫ 25 dt = 25t − 125 0 10 20 [secs] t From the nature of the problem at t = 10 seconds we may determine that the CONSTANT OF INTEGRATION is −125 meters...
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This note was uploaded on 11/29/2012 for the course PHYSICS 105 taught by Professor Tamerdoğan during the Fall '09 term at Middle East Technical University.

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