2D Motion: Vector Algebra, Displacement, Velocity, Acceleration, Projectile Motion, Slides of Physics

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September 22, 2008 Motion in Two Dimensions  Go over vector and vector algebra  Displacement and position in 2-D  Average and instantaneous velocity in 2-D  Average and instantaneous acceleration in 2-D  Projectile motion  Uniform circle motion  Relative velocity* Docsity.com September 22, 2008 Vector and its components  The components are the legs of the right triangle whose hypotenuse is A yx AAA   2 2 1tan yx y x A A A A and A            )sin( )cos(        AA AA y x Or,                      x y x y yx A A A A AAA 1 22 tanor tan   Docsity.com September 22, 2008  Average velocity  Instantaneous velocity  v is tangent to the path in x-y graph; Average & Instantaneous Velocity dt rd t rvv tavg        00t limlim jvivj t yi t xv yavgxavgavg ˆˆˆˆ ,,        t rvavg      jvivj dt dyi dt dx dt rdv yx ˆˆˆˆ    Docsity.com September 22, 2008 Motion of a Turtle A turtle starts at the origin and moves with the speed of v0=10 cm/s in the direction of 25° to the horizontal. (a) Find the coordinates of a turtle 10 seconds later. (b) How far did the turtle walk in 10 seconds? Docsity.com September 22, 2008 Motion of a Turtle Notice, you can solve the equations independently for the horizontal (x) and vertical (y) components of motion and then combine them! yx vvv  0 scmvv x / 06.925cos00    X components:  Y components:  Distance from the origin: cmtvx x 6.900  scmvv y / 23.425sin00   cmtvy y 3.420  cm 0.10022  yxd Docsity.com September 22, 2008 Motion in two dimensions tavv   0  Motions in three dimensions are independent components  Constant acceleration equations  Constant acceleration equations hold in each dimension  t = 0 beginning of the process;  where ax and ay are constant;  Initial velocity initial displacement ; 2 2 1 0 tatvrr   tavv yyy  0 2 2 1 00 tatvyy yy  )(2 0 2 0 2 yyavv yyy  tavv xxx  0 2 2 1 00 tatvxx xx  )(2 0 2 0 2 xxavv xxx  jaiaa yx ˆˆ   jvivv yx ˆˆ 000   jyixr ˆˆ 000   Docsity.com September 22, 2008  Define coordinate system. Make sketch showing axes, origin.  List known quantities. Find v0x, v0y, ax, ay, etc. Show initial conditions on sketch.  List equations of motion to see which ones to use.  Time t is the same for x and y directions. x0 = x(t = 0), y0 = y(t = 0), v0x = vx(t = 0), v0y = vy(t = 0).  Have an axis point along the direction of a if it is constant. Hints for solving problems tavv yyy  0 2 2 1 00 tatvyy yy  )(2 0 2 0 2 yyavv yyy  tavv xxx  0 2 2 1 00 tatvxx xx  )(2 0 2 0 2 xxavv xxx  Docsity.com September 22, 2008  2-D problem and define a coordinate system: x- horizontal, y- vertical (up +)  Try to pick x0 = 0, y0 = 0 at t = 0  Horizontal motion + Vertical motion  Horizontal: ax = 0 , constant velocity motion  Vertical: ay = -g = -9.8 m/s2, v0y = 0  Equations: Projectile Motion 2 2 1 gttvyy iyif  tavv yyy  0 2 2 1 00 tatvyy yy  )(2 0 2 0 2 yyavv yyy  tavv xxx  0 2 2 1 00 tatvxx xx  )(2 0 2 0 2 xxavv xxx  Horizontal Vertical Docsity.com September 22, 2008  Initial conditions (t = 0): x0 = 0, y0 = 0 v0x = v0 cosθ0 and v0y = v0 sinθ0  Horizontal motion:  Vertical motion:  Parabola;  θ0 = 0 and θ0 = 90 ? Trajectory of Projectile Motion 2 2 1 00 gttvy y  x x v xttvx 0 0 0  2 00 0 2               xx y v xg v xvy 2 0 22 0 0 cos2 tan x v gxy    Docsity.com September 22, 2008  Initial conditions (t = 0): x0 = 0, y0 = 0 v0x = v0 cosθ0 and v0x = v0 sinθ0, then What is R and h ? Horizontal Vertical 2 2 1 000 gttv y tvx x00 g v g vvtvxxR x 0 2 00000 00 2sinsincos2   g v g v t y 000 sin2 2   2 0 2 2 1 00 222       tgtvgttvyyh yhhy g vh 2 sin 0 22 0  y y yyy vg v gvgtvv 0 0 00 2  h gtvv yy  0 2 2 1 00 gttvyy y  xx vv 0 tvxx x00  Docsity.com September 22, 2008 Projectile Motion at Various Initial Angles  Complementary values of the initial angle result in the same range  The heights will be different  The maximum range occurs at a projection angle of 45o g vR 2sin 2 0 Docsity.com September 22, 2008  Centripetal acceleration  Direction: Centripetal Uniform Circular Motion r v t va r v r v t r t v r rvv r r v v r 2 2 so,                O x y ri R A B vi rf vf Δr vi vf Δv = vf - vi Docsity.com September 22, 2008 Uniform Circular Motion  Velocity:  Magnitude: constant v  The direction of the velocity is tangent to the circle  Acceleration:  Magnitude:  directed toward the center of the circle of motion  Period:  time interval required for one complete revolution of the particle r vac 2  r vac 2  v rT 2 vac   Docsity.com September 22, 2008  Position  Average velocity  Instantaneous velocity  Acceleration  are not same direction. Summary jyixtr ˆˆ)(  jaiaj dt dv i dt dv dt vd t vta yx yx t ˆˆˆˆlim)( 0        jvivj t yi t x t rv yavgxavgavg ˆˆˆˆ ,,            jvivj dt dyi dt dx dt rd t rtv yxt ˆˆˆˆlim)( 0        dt dxvx  dt dyvy  2 2 dt xd dt dva xx  2 2 dt yd dt dv a yy  )( and),( , tatv(t)r  Docsity.com