WORKSHEET 3: Vectors and Two-dimensional Motion
Problem 1: In the figure, which of the ways indicated for combining x and y components of a vector A are proper to determine the vector?
Solution:
      As discussed in class, there are two ways to graphically add or subtract vectors.  based on those methods, we see that the only correct diagrams are c, d, and f


PROBLEM SOLVING TACTIC 1: Angles – Degrees and Radians PROBLEM SOLVING TACTIC 2: Trig functions

Problem 2: (a) In the figure below what are the signs of the x and y components of both vectors? (b) How about the signs of the x and y components of d1 + d2 ?
Solution:
Recall: 1st quadrant : both x and y are positive; 2nd quadrant: x is negative and y is positive; 3rd quadrant: both x and y are negative; 4th quadrant: x is positive and y is negative.
(a) d1 is in the first quadrant therefore both x and y are positive.
To find the signs of d2 move the origin to the tail of d2.  Once you do that you notice that d2 is in the fourth quadrant therefore x is positive and y is negative.
b) By vector addition, you can see that the sum of the two is in the first quadrant therefore the x and y components are both positive.

Problem 3: In a road rally, you are given the following instructions: from the starting point use available roads to drive 36 km due east to checkpoint Able, then 42 km due north to checkpoint Baker, then 25 km northwest to Charlie.
At "Charlie" what are the magnitude and orientation of your displacement from the starting point?
Solution:
The figure shows a convenient orientation for the xy-coordinate system as well as vectors representing the three displacements they have undergone.  The scalar components of the displacement d are:
                           dx = ax + bx + cx = 36 km + 0 + (25 km)(cos 135o)
                                 = 36 km + 0 - 18 km = 18 km
                           dy = ay + by + cy = 0 + 42 km + (25 km)(sin 135o)
                                = 0 + 42 km + 18 km = 60 km

Problem 4:Consider the following vectors:
                                        A = 5.0 km due East
                                        B = 3.0 km 450 North of East
Find R = A + B using the graphical method.
SOLUTION:
Measuring the resultant R, we get 7.5 cm = 7.5 km and the angle is about 160

Problem 5: Use the same vectors in Problem 1 and solve for R using the analytical method.

SOLUTION:
Vector
x-component
y-component
A
5.0 cos0 = 5.0 km
5.0 sin0 = 0
B
3.0 cos45 = 2 .1 km
3.0 sin45 = 2 .1 km
R
7.1 km
2.1 km
                                        |R| = [(Rx)2 + (Ry)2]1/2 = 7.4 km
                                        q = tan-1(Ry/Rx) = 160


Problem 6: Southwest Airlines flight 449 flies from Lubbock to Las Vegas, then from Las Vegas to San Diego, then from San Diego to San Francisco.
                    Vector A = Lubbock to Las Vegas = 1600 km at 200 North of West
                    Vector B = Las Vegas to San Diego = 800 km at 300 South of West
                    Vector C = San Diego to San Francisco = 1000 km North
(Note: All values are fictitious)
Find the magnitude and direction of the displacement vector between Lubbock and San Francisco using the analytical method (i.e, find R = A + B + C ).
Vector 
x-component
y-component
A
  -1600 cos(20o) = -1500 km  1600 sin(20o) = 550 km
B
  -800 cos(30o) = -690 km -800 sin(30o) = -400 km 
C
  1000 cos(90) = 0  1000 sin(90) = 1000 km
R
  -2190 km  1150 km
                                          |R| = [(Rx)2 + (Ry)2]1/2 = 2,500 km
                                        q = tan-1(Ry/Rx) = -280 or 28o N of West