course Phy 231 ¯}{›¯±Üµ¸¿¬åö¥à ͵®è²Ð†ÍØassignment #010
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12:39:54 Query introductory problem set 3 #'s 7-12 Describe two ways to find the KE gain of an object of known mass under the influence of a known force acting for a given time, one way based on finding the distance the object moves and the other on the change in the velocity of the object, and explain why both approaches reach the same conclusion.
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RESPONSE --> The KE can be found by the equation KE = .5m(vf^2 -v0^2) and by the equation KE = Fnet*`ds These can be shown to be equal. Fnet = m*a vf^2=v0^2 + 2a`ds vf^2-v0^2 = 2a`ds (vf^2-v0^2)/2`ds = a Fnet = m * (vf^2-v0^2)/2`ds Fnet * `ds = .5m(vf^2 - v0^2) confidence assessment: 3
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12:40:11 ** First way: KE change is equal to the work done by the net force, which is net force * displacement, or Fnet * `ds. Second way: KE change is also equal to Kef - KE0 = .5 m vf^2 - .5 m v0^2. **
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RESPONSE --> self critique assessment: 3
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12:40:24 General College Physics and Principles of Physics: prob 2.04 convert 35 mi/hr to km/hr, m/s and ft/s.
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RESPONSE --> n/a confidence assessment: 3
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12:40:29 We need a conversions between miles and meters, km and ft, and we also need conversions between hours and seconds. We know that 1 mile is 5280 ft, and 1 hour is 3600 seconds. We also know that 1 inch is 2.54 cm, and of course 1 foot is 12 inches. 1 mile is therefore 1 mile * 5280 ft / mile = 5280 ft, 5280 ft = 5280 ft * 12 in/ft * 2.54 cm / in = 160934 cm, which is the same as 160934 cm * 1 m / (100 cm) = 1609.34 m, which in turn is the same as 1609.34 m * 1 km / (1000 m) = 1.60934 km. Thus 35 mi / hr = 35 mi / hr * (1.60934 km / 1 mi) = 56 (mi * km / (mi * hr) ) = 56 (mi / mi) * (km / hr) = 56 km / hr. We can in turn convert this result to m / s: 56 km/hr * (1000 m / km) * (1 hr / 3600 sec) = 15.6 (km * m * hr) / (hr * km * sec) = 15.6 (km / km) * (hr / hr) * (m / s) = 15.6 m/s. The original 35 mi/hr can be converted directly to ft / sec: 35 mi/hr * ( 5280 ft / mi) * ( 1 hr / 3600 sec) = 53.33 ft/sec.
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RESPONSE --> self critique assessment: 3
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12:40:41 Gen phy and prin phy prob 2.16: sports car 95 km/h stops in 6.2 s; find acceleration
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RESPONSE --> n/a confidence assessment: 3
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12:40:46 ** 95 km/hr = 95,000 m / (3600 sec) = 26.3 m/s. So change in velocity is `dv = 0 m/s - 26.3 m/s = -26.3 m/s. Average acceleration is aAve = `dv / `dt = -26.3 m/s / (6.2 s) = -4. m/s. So the time to come to a stop is `dt = `ds / vAve = 50 m / (12.5 m/s) = 4 s. Acceleration is rate of velocity change = change in velocity / change in clock time = -25 m/s / (4 s) = -4.2 m/s^2. Extension: One 'g' is the acceleration of gravity, 9.8 m/s^2. So the given acceleration is -4.2m/s^2 / [ (9.8 m/s^2) / 'g' ] = -.43 'g'.
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RESPONSE --> self critique assessment: 3
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14:10:53 univ phy 2.66 train 25m/s 200 m behind 15 m/s train, accel at -.1 m/s^2. Will the trains collide and if so where? Describe your graph.
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RESPONSE --> Yes there will be a collision. Using the equation `ds = v0`dt + .5a`dt^2 for each train, Their positions can be found at various times. The equations used were: For the train traveling 25m/s initially: 25m/s`dt + .5(-0.1m/s^2)`dt For the 2nd train traveling 15m/s: 15m/s`dt + .5(0m/s^2)`dt + 200m I plugged in times into each of these to obtain their positions. The graph of the 2nd train traveling at 15m/s was linear because there was no acceleration. It mas a line from (0,200) to (23,545) The slope of the line is of course 15m/s The first train traveling at 25m/s initially the slope of the line increased at a decreasing rate. The graphs intersected near t=23s confidence assessment: 3
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14:19:09 ** If we assume the passenger train is at position x = 0 at clock time t = 0 we conclude that the position function is x(t) = x0 + v0 t + .5 a t^2; in this case a = -.1 m/s&2 and x0 was chosen to be 0 so we have x(t) = 25 m/s * t + .5 * (-.1m/s^2) * t^2 = 25 m/s * t - .05 m/s^2 * t^2. To distinguish the two trains we'll rename this function x1(t) so that x1(t) = 25 m/s * t - .05 m/s^2 * t^2. At t = 0 the freight train, which does not change speed so has acceleration 0 and constant velocity 15 m/s, is 200 m ahead of the passenger train, so the position function for the freight train is x2(t) = 200 m + 15 m/s * t . The positions will be equal if x1 = x2, which will occur at any clock time t which solves the equation 25 t - .05 t^2 = 200 + 15 t(units are suppressed here but we see from the units of the original functions that solutions t will be in seconds). Rearranging the equation we have -.05 t^2 + 10 t - 200 = 0. The quadratic formula tells us that solutions are t = [ - 10 +- sqrt( 10^2 - 4 * (-.05) * (-200) ) ] / ( 2 * .05 ) Simplifying we get solutions t = 22.54 and t = 177.46. At t = 22.54 seconds the trains will collide. Had the trains been traveling on parallel tracks this would be the instant at which the first train overtakes the second. t = 177.46 sec would be the instant at which the second train again pulled ahead of the slowing first train. However since the trains are on the same track, the accelerations of both trains will presumably change at the instant of collision and the t = 177.46 sec solution will not apply. GOOD STUDENT SOLUTION: for the two trains to colide, the 25 m/s train must have a greater velocity than the 15 m/s train. So I can use Vf = V0 + a('dt). 15 = 25 + (-.1)('dt) -10 = -.('dt) 'dt = 100 so unless the displacement of the 25 m/s train is greater than the 15 m/s train in 100 s, their will be no colision. 'ds = 15 m/s(100) + 200 m 'ds = 1700 m 'ds = 25 m/s(100) + .5(-.1)(100^2) = 2000 m. The trains collide. **
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RESPONSE --> I had used the quadratic formula to find the time of impact but left it off and I had also found out how much time it would have taken for the first train to slow to 15m/s in order to avoid a collision. self critique assessment: 2
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