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Unit 2 motion in 1-dim_A_CT lec 1 part2

Unit 2 motion in 1-dim_A_CT lec 1 part2 - Unit 1 Motion in...

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1 Unit 1 Motion in 1-dim Part 2 2 Graphical Analysis of 1-dim motion Motion with Constant Acceleration Solving Problems with a=const Freely Falling Objects Presentation Goals CT #1: A particle is moving in a straight line that its position is given by the equation x(t) = 2.10 t 2 + 2.80 The particle’s average acceleration during the time interval t 1 =3.0 s and t 2 =5.00 s is 2.10 m/s2 2.80 m/s2 4.20 m/s2 5.00 m/s2 0% 0% 0% 0% Graphical Analysis of 1-dim 1. 2.10 m/s 2 2. 2.80 m/s 2 3. 4.20 m/s 2 4. 5.00 m/s 2 0 of 200 10 CT #2: A particle is moving in a straight line that its position is given by the equation x(t) =2.10 t 2 +2.80 The particle’s instantaneous acceleration at t=5.00 s is 2.10 m/s2 2.80 m/s2 4.20 m/s2 5.00 m/s2 0% 0% 0% 0% Graphical Analysis of 1-dim 1. 2.10 m/s 2 2. 2.80 m/s 2 3. 4.20 m/s 2 4. 5.00 m/s 2 0 of 200 10 5 Q: Can we construct an analytical model of ANY motion with a=const ? Q#1: What is the velocity vs. time? A: for any interval t =t-t 0 =t -0 dv/dt = const = a, so Now, we can solve the velocity vs. time : t a + = 0 v v Constant Acceleration (A) at v v dt a dv v v t = = 0 0 0 6 Q: How to represent graphically this model? Constant Acceleration A: Its velocity vs time graph is a straight line, but not necessarily through the origin Q#2: How the position of the particle varies with time when we know a=const and the initial conditions x 0 and v 0

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7 Fact: For any motion with t 0 =0: In addition, as the velocity is increasing at a constant rate , the average velocity is arithmetic average of the velocity at start and at the end: we can eliminate in Eq. 2-2 and get the displacement vs. time: (B) t x t t x 0 0 0 x x v = = v ) v v ( 2 1 v 0 + = t x x ) v v ( 2 1 0 0 + = Constant Acceleration (2-2) (2-9) 8 Eliminating the final velocity from v=v 0 +at we find another form of the displacement vs. time: (C) 2 0 0 2 1 v at t x x + = t x x ) v v ( 2 1 0 0 + = Constant Acceleration or eliminating the initial velocity from v 0 =v-at we find the third form: 2 0 2 1 v at t x x = (D) (B) 9 (E) Q#3: How to calculate v
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Unit 2 motion in 1-dim_A_CT lec 1 part2 - Unit 1 Motion in...

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