9-8-08LectPHY2048 - PHY2048 9/8/08 Position, velocity,...

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Position, velocity, acceleration vs. time plots (demo). What causes acceleration? Projectile motion PHY2048 9/8/08
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time (s) position (m) time (s) velocity (m/s) time (s) accel. (m/s 2 ) time (s) time (s) time (s) dx v (constant) dt == o v(t) v (constant) = = dv a0 dt = = O x Object at rest (wrt coord. syst.): Constant velocity (wrt coord. syst.): 0 0 0 0 0 0 O x(t) x constant O x O x vt = +
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time (s) position (m) time (s) velocity (m/s) time (s) accel. (m/s 2 ) dx(t) v(t) (not constant) dt == dv(t) a(t) dt = ac o n s t a n t = 2 2 d x(t) dt dt Note that: 2 OO x(t) x v t t = ++ α This means that only a quadratic (and lower) time dependence of x(t) can give a constant acceleration. O dx vt dt = 2 2 dx a 2 (time independent) dt α O x O v E.g. suppose x(t) Cart Motion Sensor demo 0 0 0 Constant acceleration has a parabolic form 2 1 2 x at = O v =+
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3-Dimensional Kinematics 1-Dimension 3-Dimensions ˆ x(t) x(t)i = G dx(t) ˆ v(t) i dt = G dv(t) ˆ a(t) i dt = G position: velocity: xyz ˆˆˆ r(t) r (t)i r (t)j r(t)k =+ + G acceleration: y xz dr (t) i j k dt dt dt =++ G y dv (t) i j k dt dt dt G v (t)i v (t)j v( t )k G a (t)i a (t)j a( t G
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where: r G ˆ i ˆ j ˆ k x r z r y r
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HITT#1 (9/8/08) 1 r G 2 r G & are the position vectors of an object at The displacement vector of the object during this time is: 12 rrr Δ= + G GG A. B. C. D. E. times t 1 and t 2 (t 2 later than t 1 ). Δ= − G 21 GGG rr r Δ= ⋅ G Δ= × G Hint: Recall position vectors are always from the coordinate system origin to the object. Displacement vectors are from the objects initial position to the later position. Draw the 3 vectors.
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1 r G 2 r G r Δ G By the head to tail rules of geometric vector addition clearly, Hence 12 rrr Δ= + G G A. B. C. D. E. Δ= − G G 21 rr r G G r Δ= ⋅ G G Δ= × G G Δ =− G GG r + Δ= G (at t 1 ) (at t 2 ) By definition the displacement vector points from the earlier to the later position vectors.
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1 ˆˆ r2 i 2 j = + G 2 r5 i 2 j = G HITT#2 (9/8/08) These position vectors are (in meters): So the magnitude of is: 21 rrr Δ =− G GG m Δ = A. B. C. D. E. r8 m Δ = r6 m Δ = . 6 m Δ = m Δ =
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r5 m Δ= A. B. C. D. E. r8 m r6 m r2 . 6 m m 1 ˆˆ i 2 j = + G 2 i 2 j = G 21 rrr Δ =− G GG r3 i 4 j Δ G 22 2 2 r( 3 m ) ( 4 m ) 9 m 1 6 m +− = + 2 r 25m 5m =
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In 1-D kinematics we derived that for constant acceleration 2 1 oo 2 xx v t a t =+ + o vv a t = + It can similarly be shown that in 3 dimensions, 2 1 2 rr v t a t + GG G G o a t = + G G G Where now each of these vector equations is really 3 equations, one for the components along each coordinate direction: 2 1 o x o x 2 v t a t + 2 1 yy o y o y 2 v t a t + 2 1 zz o z o z 2 v t a t + o x a t = + o y a t = + o z a t = + x: y: z:
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Also, in 1-D we developed the expression, ( ) 22 oo vv 2 a x x = +− In 3-D we have this becomes 3 equations, ( ) xx o x x x o 2 a r r = ( ) yy o y y y o 2 a r r =+ ( ) zz o z z z o 2 a r r = These and the foregoing equations will let us handle (predict) x:
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9-8-08LectPHY2048 - PHY2048 9/8/08 Position, velocity,...

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