1. Principle of superposition: If an object is subjected to two separate influences, each producing a characteristic type of motion, it responds to each without modifying its response to the other.
   
   
2. A vector analysis allows you to separate two-dimensional motion into two one-dimensional motions and then combine them at the end of the problem.  Example: projectile motion
   
   
   

1. The polygon method.  In Fig. 1a below, the vectors are placed head to tail.  The resultant vector goes from the tail of the first vector A to the head of the second vector B to give C = A + B
   
   
   
2. The parallelogram method.  Draw the vectors with their tails at one point.  Complete the parallelogram and draw the diagonal to find the resultant. See Fig. 1b below.
   
   
   

1. Unit vectors.  The unit vector i, j, and k are unit vectors along the X, Y, and Z-axes, respectively.  When multiplied by a number or a symbol that represents a quantity it becomes a vector with the magnitude of the quantity (or symbol).  For example, 5m i is a vector of length 5 m along the X-axis.
   
   
2. Components of vectors
   
   
   1. If a vector lies in the X-Y plane, it can be written as a component in the X-direction added vectorally to a component in the Y-direction. A = A_xi + A_yj, where A_x and A_y are the X and Y components of the vector A (Fig. 2a below).
      
      
      
      
   2. From Fig. 2b above, we see that A_x = A cos Θ A_y = A sin Θ
      and A = (A_x²  + A_y²)^1/2
      
      
   3. Utility of components.  If two vectors are equal, their components along any chosen axis are equal.   If C = D,  then C_x  = D_x
      and C_y  = D_y
      
       
   4. Example (Fig. 3 below)
      
      
      
      A = A_xi + A_xj     B = B_xi + B_xj
      A + B = (A_xi + A_yj) + (B_xi + B_yj)
                = (A_x + B_x))i + (A_y+ B_y)j
      C = C_xi + C_xj.
        If C =A + B, C_x =(A_x + B_x)), and
      C_y  =(A_y+ B_y) [Fig. 3 above]
      
      

1. In the X-direction

   v_x = v_ox + a_xt   and  x = x_o + v_ox t + 1/2 a_xt²
   

   
   
2. In the Y-direction

   v_y = v_oy + a_yt   and  y = y_o + v_oy t + 1/2 a_yt²
   

   
   
   
   
     

v = v_o + at   and  r = r_o  + v_ot + 1/2 at²,
where r_o  = x_o i + y_oj, v_o  = v_oxi  + v_oy j, and a = (a_xi + a_yj)
r  = xi + yj.  From A we see we may write:
r = (x_o + v_ox t + 1/2 a_xt²)i + (y_o + v_oy t + 1/2 a_yt²)j,  or
r = (x_oi  + y_oj) + (v_oxi  + v_oy j)t + 1/2(a_xi + a_yj)t^2,  or
r =      r_o             +       v_ot           +       1/2 at²

1. Sample problem.   An object is given an initial velocity of 25 m/s at an angle of 53⁰ with the horizontal.  Find the initial position of the object along (a) the X-axis, x_o,  (b) the Y axis, y_o,  the initial velocities along  (c) the X-axis, v_ox, and (d) the Y-axis, v_oy.  (e) Indicate the directions of v_ox and v_oy on the figure. Find (f) the maximum vertical height y_max, (g) the time for object to hit the ground, (h) the distance x the projectile travels in a horizontal direction, (i) v_x, v_y, and v just before the object hits the ground, and (j) the angle v makes with the horizontal. Take g = 10 m/s².
   
   
   
   
2. Approach to solution
   
   
   1. Choose an appropriate X and Y axis.  For a projectile problem, the appropriate X-axis is the horizontal direction and the Y-axis the vertical direction.  Draw the figure, as shown in Fig. 4b below.
      
      
      
      
   2. Set up a chart for the X and Y components of the descriptive quantities. Identify the quantities asked for in (a) through (d) of the statement of the problem along with the components of the acceleration and enter them in the chart. Then write appropriate formulae for the components of motion.
      
      

      X	Y
      (a) xo = 0	(b) yo =25 m
      (c) vox = 25 m/s cos 53o   =15 m/s	(d) voy = 25 m/s sin 53o    = 20 m/s
      ax = 0	ay = -10 m/s2
      x(t) = xo + voxt + 1/2 axt2	vy2(y) = (voy)2 + 2ay(y - yo)
      (e) x(t) = 0 + 15 m/s t + 0	(f) At ymax, vy2(y) = 0 = (20m/s)2- 20m/s2(ymax-25m) (ymax- 25 m) = 20 m,  or ymax = 45 m
      	(g) y(t) = yo + voyt + 1/2 ayt2 when object hits ground, 0 = 25 m + 20 m/s t - 5 m/s2 t2, or  t2- 4 s t - 5 s2 = 0 (t - 5 s)(t +1 s) = 0 t = 5 s.  t cannot be negative.
      (h) x(5s) = (15m/s)(5s) = 75m
      (i) vx(t) = vox + axt vx(5 s) = vox = 15m/s	(i) vy(t) = voy + ayt vy(5 s) = 20 m/s -(10 m/s)(5s) = -30 m/s
      v = (vx2 + vy2)1/2 = [152 + (-30)2]1/2 m/s = 33.5m/s
      (j) tan Θ = vy/vx = -30/15 = - 2.            Θ = - 63o

      
      
      

1. Instantaneous velocity tangent to the circle at any instant.
   
   
2. Displacement vector sweeps through same angle as the velocity vector.
   
   
   

1. Look at the triangles of Fig. 6b above.
   
   
2. Δv/v = Δr/r  or  Δv = v Δr/v.
   
   
3. To find magnitude of acceleration consider:

   Δv/ Δt = (v/r)( Δr/ Δt)
   
   a = lim Δv/ Δt = (v/r)(v) = v²/r
         ^{Δt o}
   

   
   
4. Since a = lim of Δv/ Δt as Δt 0, find the direction of the acceleration by looking at the direction of Δv as Δt approaches 0.   Δv and a point into the center of the circle.
   
   
   

1. The angle ΔΘ subtends an arc whose length is Δs.
   
   
2. By definition, the angle in radians,  ΔΘ= Δs/r.
   
   
3. Since v =  Δs/ Δt,  ΔΘ/ Δt = ( Δs/r)/ Δt = v/r.  The limit as Δt approaches zero of ΔΘ/ Δt is ω.  So ω = v/r.
   
   
4. The direction of ω is given by the right hand rule:  When the fingers of your right hand are curled in the direction of the motion, the direction of your thumb points in the direction of ω.  For Fig. 6a above, ω is out of the page.
   
   
5. While v is not constant for uniform circular motion,  ω  is constant.
   Θ = ωt.