Lecture3PHY2048

# Lecture3PHY2048 - Summarizing the 1-d kinematic formalism,...

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Summarizing the 1-d kinematic formalism , we have, ˆ x(t) x(t)i , = G dx(t) v(t) dt = G G dv(t) a(t) dt = G G & Lovely Position vector Velocity Acceleration ..., but what use is that? Where we show the (possible) time dependence explicitly. Well from these we can obtain predictive expressions.

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From the expression for acceleration : dv a dt = adt dv = dv adt = o vt t v0 0 dv adt a dt == ∫∫ G G if is constant a G o (v v ,at t 0) = = G G then o vv a t = o a t = + o , a t t i m e t0 Hence if the acceleration is constant & then at later time t, The fundamental theorem of integral calculus lets us integrate both sides (switch sides)
Now from the expression for velocity we have that: dx v dt = Using this in the expression we just derived ( ), o vv a t = + o dx va t dt = + G G Multiplying through by dt, o dx (v at)dt = + Now integrating this, o xt o x0 dx =+ ∫∫ G G

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tt oo 00 xx v d t a t d t −= + (since constant) o v,a G G so, 2 1 2 v t a t =+ + v d ta t d t + t 2 0 2 1 t 2 1 t 2 ⎡⎤ ⎢⎥ ⎣⎦ = o vv a t 2 1 2 v t a t + Hence we’ve derived the kinematic equations, which for known constant acceleration let us predict the future position and velocity given the initial position, velocity (at t = 0) and constant acceleration.
There are other equations derived from these that are sometimes useful. 2 1 oo 2 xx v t a t =+ + o vv a t For example we can use these to derive an expression that does not require knowledge of the time. To do this solve the equation on the right for the time: o a t −= o a t = + o t a = o t a = ( ) 2 o 2 2 t a = & We can now use these in the eqn on the left above eliminating the time, ( ) 2 o o 1 2 2 v a aa ⎛⎞ + ⎜⎟ ⎝⎠

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( ) 2 2 o oo 1 o 2 vv vv v xx aa −= + ( ) 2 o o 1 2 2 v a ⎛⎞ + ⎜⎟ ⎝⎠ Simplifying, ( ) ( ) 2 2 1 ooo o 2 ax x v v + ( ) ( ) 22 2 1 o o 2 v v 2 v v + + ( ) 222 11 v v + + ( ) v v ( ) 2a x x v v
( ) 22 oo vv 2 a x x = +− ( ) 2a x x v v =− ( ) v2 a x x v + −= For the final result that, Where v o is the velocity at the position x o .

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o vv a t = + G GG 2 1 oo 2 xx v t a t = ++ G G As stated in HRW, other expressions can be formulated, in which different variables have been eliminated (Table 2-1, p.23). We repeat here for completeness. Eqn # Expression Missing parameter 2-11 2-16 2-15 2-17 2-18 o v G 22 2 a ( x x ) = +− t 1 2 ( v v ) t = a 2 1 o 2 v t a t = o v memorize Caveat: valid only for constant acceleration
What these mean (example): A sprinter runs passed her coach at a constant velocity of 8 m/s when the coach starts his stop watch . How far is she from the coach when the stop watch reads 11 s ? Obviously (8m/s)(11s) = 88 m.

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## This note was uploaded on 08/25/2011 for the course PHY 2048 taught by Professor Field during the Fall '08 term at University of Florida.

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Lecture3PHY2048 - Summarizing the 1-d kinematic formalism,...

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