Motion in One Dimension
Motion in One Dimension PHY 2048C
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This 3 page Class Notes was uploaded by Harrison Harward on Monday January 25, 2016. The Class Notes belongs to PHY 2048C at Florida State University taught by H. Prosper in Winter 2016. Since its upload, it has received 106 views. For similar materials see General Physics in Physics 2 at Florida State University.
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Date Created: 01/25/16
11416 PHY 2048C Motion in One Dimension Class problem w acceleration Motion with Acceleration Constant Acceleration 0 If acceleration is constant the slope is the same o It is rigorously true a mathematical fact vvo area under at curve at xxo area under vt curve vot12 at2 o What we have discovered here is integration or calculus xxo vot12at2 Equations of motion for constant acceleration 0 One can easily eliminate either a t or vo by solving equations 27 and 210 simultaneously 27 vvoat 210 x x0 vot 12at 211 v vo 2a xxo 29 xxo l2vovt 292 xxo vt 12at Example 1D Projectile Motion 0 A ball is thrown upwards at 5ms relative to the ground from a height of 10m above the ground Neglect air resistance We need to choose a coordinate system We shall choose y0 to be ground level and up to be the positive y direction Speed and velocity decrease in the upward direction Acceeration is down Speed increases in the downward direction velocity becomes more negative increases negatively T v0 5 ms y 10m 1 How high above the 1 39 th ground WI e ba V2 V02 29y yo reach We know 0 V02 3929y yo 2 2 V0 y yO VOZg V0 5 mS 2 Ytop 10m 5 ms2 2 X A g 10 ms 10mSZ We eed tquot W Y nal delta y yyo vo a v O top 2 How long does it take the 2 ball to reach the ground y yovot 12gt2 We need to know t when y O O 10m 5ms t 12 We know y yo v0 a 10m52t2 We need to know t 10 5t 5t2 yyovot12lt31t2 5t2 5t 1OO O lOm5mst 12 5t5 t20 10m52t2 10 5t 5t2 5t2 5t 1OO 5t 5 t 2 0 3 At what speed does the ball t 1 S or t 25 hit the ground v vo gt 5ms 10ms2 25 5 20ms The car is dropped from some unknown height Habove the top of the eld of view say a window and takes some unknown time t1 to reach the top of the window We are told that the car must cross the window of known height h in a given that is known time t The task is to nd an expression for H in terms of known quantities Develop Since this is a motion problem we need to choose a coordinate system X0 0 to be the car39s initial position There are 2 unknowns the one we care about H and the one we don39t t1 So we need 2 equations From what we have so far we can write x xovot112gt12 H 00129t12 Hl29t12 H h 129 t1t2 l2 gt12h 12 gt1 t2 t1 t2 t12 2hgt 2t1t hgt t1 t2 t1 hgt t2 pug this back in at top H129t12
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