chapter2.5-2.6_slides

chapter2.5-2.6_slides - 9/10/2009 Constant acceleration...

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9/10/2009 1 Constant acceleration motion: Average acceleration = instantaneous acceleration , i.e., Velocity should change uniformly (linearly) with time. So we expect: a a at v t v 0 ) ( v(t): Velocity at time t A car starts with a velocity v 0 =2.0 m/s to the right and accelerates to the right with a constant a=3.0 m/s 2 . What is the car’s velocity at t = 10 sec? The driver then slammed on the break to avoid an accident. The car needs to come to a complete stop within 3 seconds. Find the acceleration. at v v 0 at v t v 0 ) s 10 ( m/s 32 0 . 10 0 . 3 0 . 2 mph 72 about at v t v 0 ) 3 ( 3 32 0 a 2 m/s 67 . 10 a
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9/10/2009 2 For constant acceleration motions, we can calculate the average velocity easily: Just the average of the initial and final velocity From the definition of average velocity: we have 2 0 v v v t v t v x t v v t v x 2 0 Substitute into the above equation, we have t v v t v x 2 0 t at v v 2 ) ( 0 0 2 0 2 1 at t v 2 0 2 1 at t v x 2 0 0 0 2 1 have we , Let at t v x x x x x at v v 0 t v v x 2 0
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9/10/2009 3 We can also an expression that does not involve time First, starting from , solve for t, we have Then, use the expression at v v 0 a v v t v v at 0 0 t v v t v x 2 0 ) )( 2 ( 0 0 a v v v v x 2 0 2 2 v v x a a 2 by sides both multiply 2 2 ) )( ( recall b a b a b a or x a v v 2 2 0 2 Summary at v v 0 2 0 2 1 at t v x 2 0 0 2 1 at t v x x 2 0 2 2 v v x a
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chapter2.5-2.6_slides - 9/10/2009 Constant acceleration...

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