Transcription of AP Physics Chapter 2 Review - cf.edliostatic.com
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AP Physics Chapter 2 Review - cf.edliostatic.com
AP PhysicsChapter 2 Review2 v= x t=21+66 54() 21+22 6()[]m2sec= 2msec v=dvdt=d21+22t 6t2()dt=22 12tv=v1sec+v3sec2=10 14()msec2= position of a particle moving along the x axis is given by x = (21 + 22t - t2) m, where t is in seconds. What is the average velocity during the time interval t = sec to t = sec?3 vf2=vi2+2a xvf2 vi22 x=a220msec()2 450msec() ()=a 550357msec2= bullet is fired through a board, cm thick, with its line of motion perpendicular to the face of the board. If it enters with a speed of 450 msec and emerges with a speed of 220 msec, what is the bullet's acceleration as it passes through the board?4 v=dvdt=12t 3t2a=dvdt=12 6tAt max v, set a = 0.
4 € v= dv dt =12t−3t2 a= dv dt =12−6t At max v, set a = 0. t=2sec x=6(2)2 −23 x=16m 3. The position of a particle moving along the x axis is given by x = 6.0t2 - 1.0t3, where x is in meters and t in seconds. What is the position of the particle when it …
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