5lecture2a08

5lecture2a08 - Physics 2A Lecture 5: Jan 16 Vivek Sharma...

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Physics 2A Lecture 5: Jan 16 Vivek Sharma UCSD Physics
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Announcements • Be sure to carefully study the the posted quiz2 solution and compare them with your own answer. • PB session for the remaining weeks are – Section A: Monday, 7pm CSB002 – Section B: Monday, 5pm CNTR119 – Section C: Monday, 6pm WLH2001 • 6pm PB session has the largest room, go there • No PB next Monday, Friday DI Æ PB session •C h e c k o u t http://solved.ucsd.edu/week1/ for streaming videos of problem solving methods
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Evolution of x vs t when a x =Constant 10 x 1 0 x 2x 2 x 0 av-x At time t 0, object at x = x ,has v v At time t t, object at x = x,has v v xx then v t = = == = x 0x x When a constant, velocity changes at const rate vv so in time interval [0 t], v 2 = + →= () x0 x av-x 0 x x x x 1 2 1 a t Bu v a t at = v 2 t since =+ + +
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Evolution of x vs t when a x =Constant 0 av-x xx v t = 0x x 1 and v t v 2 a = + x 2 0 x 0 x t x Parabolic curve in x x v 1 at 2 1 x t v a 2 t t - + = + = +
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Relating x, v x & a x (without time t) x0 x x x x vv Since a = , write t = ta 0 2 x x 0x 2 x 00 x x xx substitute for t in x 1 x=x 1 at 2 v x a 2a t v a =+ + ⎛⎞ −− ⇒+ + ⎜⎟ ⎝⎠ 22 0 2 0 x x x x x x x 0 (x x )2a 2v v 2 v 2 a ( ) ⇒− = + ++ =
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An Expression Without a x 0x x 0 a v-x a xx Since v a v nd t v v 2 = + = 0 x vv 2 t + = 0 x 0 This is a useful expression to have when a constant but unknown t 2 = + ⎛⎞ ⎜⎟ ⎝⎠ −=
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Equations Of Motion For Object under Constant Acceleration x0 x x vv a t =+ 0 2 x 0 x vt 1 at 2 x x + 22 x x 0 2 a ( x x ) 0x x 0 2 x t x + = ⎜⎟ ⎝⎠
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2 Motorcyclist going east, accelerates after passing signpost. He accelerates at 4.0m/s .At t=0, he is 5.0m east of signpost, moving east 15 m/s. (a) find his position and velocity at t=2.0s. Where is motorcyclist when his velocity is 25 m/s ? 2 00 x x xx Take signpost as origin of coordinate (x=0), East +x At t=0,x =5.0m,v =15m/s; a =4.0m/s (a) what is x,v at t=2.0s, (b) x when v =25m/s 2 x x 22 x0 x x 1 (a) Use x = x v t a t 2 1 5.0m (15m/s)(2.0s) (4.0m/s )(2.0s) 43m 2 Velocity at x=43m: v v a t 23m/s ++ =+ + = = x x 0 2 2 x 0 vv 2 a ( x x ) v v (25m/s) (15m/s) x = (b) s no t given ! o use x 5.0m 55m 2a 2a −− ⇒+ = + =
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Motion With Constant Acceleration: Freely Falling Bodies Aristote (4 BC) believed believed (didn’t check!)
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5lecture2a08 - Physics 2A Lecture 5: Jan 16 Vivek Sharma...

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