Problem 3, 4, 5

These problems involve graphically drawing vectors. I will not post solutions but I will mention that you have to do the following things:

  1. Pick an appropriate scale (example, let 1 cm = 1 m, so a 3.0 m vector will be 3.0-cm long in your solutions).
  2. You need a ruler to measure the lengths and a protractor to measure angles. Then draw an appropriate coordinate axis.
  3. Draw the vectors tip to tail then draw the resultant vector like I showed you in class.
  4. Measure the length of the resultant vector and re-scale it using your chosen scale. Then measure the angle of the resultant vector.
Problem 8

A person walks 25.00 north of east for 3.10 km. How far would a person walk due north then due east to arrive at the same location?
SOLUTION:
Basically, this question is asking you to find the components of a vector 25.00 north of east with a magnitude of 3.10 km.
x-component = (3.10 km)(cos 25.00) = 2.81 km
y-component = (3.10 km)(sin 25.00) = 1.31 km


Problem 9
A girl delivering newspapers covers her route by going 3.00 blocks west, 4.00 blocks north, then 6.00 blocks east.
(a) What is her resultant displacement?
SOLUTION:
Vector 1
Magnitude = 3.00 blocks
Direction = west or q = 00
x-component = (3.00) cos 0 = 3.00
y-component = (3.00) sin 0 = 0

Vector 2
Magnitude = 4.00 blocks
Direction = north, or q = 900
x-component = (4.00) cos 90 = 0
y-component = (4.00) sin 90 = 4.00

Vector 3
Magnitude = 6.00 blocks
Direction = east or q = 1800
x-component = (6.00) cos 180 = -6.00
y-component = (6.00) sin 180 = 0

RESULTANT VECTOR
x-component = 3.00 + 0 –6.00 = -3.00 blocks
y-component = 0 + 4.00 + 0 = 4.00 blocks
Resultant Displacement = [(-3.00)2 + (4.00)2]1/2 = 5.00 blocks
q = tan-1[4.00/(-3.00)] = -53.00 or 1270 or 53.00 North of East.

Problem 13
A commuter flight flies from an airport to city A, then to city B, then to city C.
                    Vector A = airport to city A = 175 km at 30.00 North of East
                    Vector B = Las Vegas to San Diego = 150 km at 20.00 West of North
                    Vector C = San Diego to San Francisco = 190 km West
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 ).  Remember, I will measure every angle relative to the positive x-axis or East.
Vector 
x-component
y-component
A
  175 cos(30.0o) = 151.6 km  175 sin(30.0o) = 87.5 km
B
  150 cos(110o) = -51.3 km 150 sin(110o) = 141 km 
C
  190 cos(180o) = -190  190 sin(180o) = 0
R
  -89.7 km  228.5 km
                                          |R| = [(Rx)2 + (Ry)2]1/2 = 245 km
                                        q = tan-1(Ry/Rx) = -68.60 or 111.4 or 68.6o N of West