9-15-08LectPHY2048

9-15-08LectPHY2048 - Summary of 3d kinematics An object...

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Summary of 3d kinematics x z y An object travels through 3d-space in time. r(t) G X r( t) Z Y t ) XYZ ˆˆˆ r (t)i r (t)j r(t)k =++ G It has position vector, The objects position is projected onto each coordinate axis resulting in three coupled 1d equations. These each give the position along the 1d coordinate as a function of time.
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Consider the r x (t) component. We plot it as a function of time and look at the object’s position at two times. r x t r xi t i The average speed between those times is, Xf Xi X X(Ave.) fi rr r v tt t Δ == Δ If we take the limit as t 0 Δ= − → r xf t f t Δ v X (t) XX t0 X rd r Lim v (t) td t Δ→ Δ Δ We get the instantaneous speed , v X (t) , in the x direction at the time t. v Ave
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We can similarly plot the instantaneous speed component v X as a function of time and similarly define the average x acceleration between those times as, fi tt t 0 Δ =− Again taking the limit as Xf Xi X X(Ave.) vv v a t Δ == Δ XX t0 X vd v Lim a (t) td t Δ→ Δ Δ We get the instantaneous acceleration at the time t. With similar definitions for v Y , a Y , v Z and a Z .
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Using the fundamental theorem of integral calculus, we derived for the r X component the expressions (for constant acceleration only ): 2 1 XX O X O X 2 r( t ) r v t at =+ + O X v( t ) v = + & And upon eliminating time from these: ( ) 22 O X X X O vv 2 a r r = +− With similar equations for y & z: 2 1 YY O Y O Y 2 t + O Y t ( ) O Y Y Y O 2 a r r 2 1 ZZ O Z O Z 2 t = ++ O X t = + ( ) O Z Z Z O 2 a r r = By orienting our coordinate system properly we can minimize labor making some of these vanish.
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The requirement for constant acceleration is a rather severe constraint but consider a plot of the x component of the v-t plot under some constant acceleration and corresponding a-t plot: time v dv a dt = O v 0 Dropping the subscript & rearranging, Integrating both sides, dv adt = o vt t v0 0 dv adt a dt == ∫∫ time a 0 t o 0 vv a d ta t t Area under curve ( a ) from t = 0 to t
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But suppose now that the acceleration is not constant time v dv(t) a(t) dt = O v 0 Rearranging, Integrating both sides, dv a(t)dt = o vt v0 dv = t o 0 vv a ( t ) d t −= Now if a(t) is an integrable function (or can be approximated as such) then we do the integral to still get a closed form solution. E.g. suppose , qt = ( q a constant) Then, t 2 o 0 q v v qtdt t 2 = 2 o q t 2
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2 O q v(t) v t 2 =+ Giving Now since, X X dr v dt = dr vdt = Integrating, O rt r0 dr vdt = t 2 OO 0 q rr ( v t ) d t 2 −= + tt 2 00 q v d t t d t 2 + 3 XX O X O q r( t ) r v t t 6 + So for the time dependent position we get: where we dropped the x subscript but remember that we are along x (only).
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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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9-15-08LectPHY2048 - Summary of 3d kinematics An object...

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