2. Vectors
•Vectors have magnitude (length)
and direction
•In previous chapters, we have
treated “direction” as being only
positive or negative
•Vectors are represented by
arrows; length of the arrow
corresponds to magnitude
(length) of the vector
8. ResolvingVectors (usingTrig)
•An ant leaves his nest and, after foraging for some time, is at the
location given by the vector r in the diagram below. This vector
has the magnitude r = 1.50 m and Θ = 25.0°.
•Resolve this vector into its components using sine and cosine.
r = 1.50 m
ry
rx
Θ = 25.0°
9. ResolvingVectors (usingTrig)
•An ant leaves his nest and, after foraging for some time, is at the
location given by the vector r in the diagram below. This vector
has the magnitude r = 1.50 m and Θ = 25.0°.
•Resolve this vector into its components using sine and cosine.
r = 1.50 m
ry
rx
Θ = 25.0°
cos 𝜃 =
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
=
𝑟𝑥
𝑟
10. ResolvingVectors (usingTrig)
•An ant leaves his nest and, after foraging for some time, is at the
location given by the vector r in the diagram below. This vector
has the magnitude r = 1.50 m and Θ = 25.0°.
•Resolve this vector into its components using sine and cosine.
r = 1.50 m
ry
rx
Θ = 25.0°
cos 𝜃 =
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
=
𝑟𝑥
𝑟
Rearrange this equation…
𝑟𝑥 = 𝑟 cos 𝜃
11. ResolvingVectors (usingTrig)
•An ant leaves his nest and, after foraging for some time, is at the
location given by the vector r in the diagram below. This vector
has the magnitude r = 1.50 m and Θ = 25.0°.
•Resolve this vector into its components using sine and cosine.
r = 1.50 m
ry
rx
Θ = 25.0°
cos 𝜃 =
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
=
𝑟𝑥
𝑟
Rearrange this equation…
𝑟𝑥 = 𝑟 cos 𝜃
𝑟𝑥 = (1.50 m) cos 25°
𝑟𝑥 = 1.50 m 0.906
𝑟𝑥 = 1.36 m
12. ResolvingVectors (usingTrig)
•An ant leaves his nest and, after foraging for some time, is at the
location given by the vector r in the diagram below. This vector
has the magnitude r = 1.50 m and Θ = 25.0°.
•Resolve this vector into its components using sine and cosine.
r = 1.50 m
ry
rx
Θ = 25.0°
sin 𝜃 =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
=
𝑟𝑦
𝑟
13. ResolvingVectors (usingTrig)
•An ant leaves his nest and, after foraging for some time, is at the
location given by the vector r in the diagram below. This vector
has the magnitude r = 1.50 m and Θ = 25.0°.
•Resolve this vector into its components using sine and cosine.
r = 1.50 m
ry
rx
Θ = 25.0°
sin 𝜃 =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
=
𝑟𝑦
𝑟
Rearrange this equation…
𝑟𝑦 = 𝑟 sin 𝜃
14. ResolvingVectors (usingTrig)
•An ant leaves his nest and, after foraging for some time, is at the
location given by the vector r in the diagram below. This vector
has the magnitude r = 1.50 m and Θ = 25.0°.
•Resolve this vector into its components using sine and cosine.
r = 1.50 m
ry
rx
Θ = 25.0°
sin 𝜃 =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
=
𝑟𝑦
𝑟
Rearrange this equation…
𝑟𝑦 = 𝑟 sin 𝜃
𝑟𝑦 = (1.50 m) sin 25°
𝑟𝑦 = 1.50 m 0.423
𝑟𝑦 = 0.634 m
19. CombiningVectors
0.60 km
0.60 km ?
To find direction, use trigonometry
tan 𝜃 =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
=
𝑟𝑦
𝑟𝑥
𝜃 = tan−1(
𝑟𝑦
𝑟𝑥
)
𝜃 = tan−1
(
0.60 km
0.60 km
)
𝜃 = 45°
20. Practice Problems
1. Resolve this vector into its horizontal (x) and vertical (y)
components: A cannonball is fired with a velocity of 34.0 m/s
at an angle of 31.5° to the horizontal.
2. Combine these two vectors:A plan flies due west at 245 m/s
and is pushed by a northward wind blowing at 29.0 m/s. What
is the resultant vector for the plane’s velocity?
Hint #1: Draw a picture for each!
Hint #2:To determine sides of a triangle use trig functions. To
determine an angle, use inverse trig functions (tan-1, etc.)
21. Challenge Problem
•Resolve each of the three vectors to answer the following
question.
•A man is exploring with a compass in the dark. He is 300 meters
west from a lake, the coastline of which extends due north and
south. He follows the following three directions. (All angles are
measured from the horizontal.)
•Walk 250 meters NE at 40 degrees.
•Walk 100 m NW at 30 degrees.
•Walk 300 m NE at 75 degrees.
•How far is he from the lake?
Lake
1
2
3
?
300 m
22. Challenge Problem
•Combine the three vectors to answer the following question.
•A man is exploring with a compass in the dark. He is 300 meters
west from a lake, the coastline of which extends due north and
south. He follows the following three directions. (All angles are
measured from the horizontal.
•Walk 250 meters NE at 40 degrees.
•Walk 100 m NW at 30 degrees.
•Walk 300 m NE at 75 degrees.
•What is his resultant vector?
Lake
1
2
3
?
24. Relative Motion
•Often, motion must be described
relative to the motion of another
object.
•We do this all the time without
realizing it… for instance, your car
drives 50 mi/hr down a highway
relative to the motion of the ground.
•Think about it: A train is moving at 15.0
m/s. A man walks towards the front of
the train at 1.2 m/s. What is his speed
relative to the ground? What if he is
walking at 1.2 m/s towards the back of
the train?
25. Relative Motion
•If the motion occurs in two dimensions, simply add the vectors
using the strategies from earlier in the chapter.
28. Projectile Motion
•An object that is
thrown will follow a
curved path (parabola)
to an onlooker.
•Air resistance must be
ignored (so horizontal
velocity is constant)
•Acceleration due to
gravity is -9.81 m/s2.
31. Practice Problems
•You throw a baseball off the roof of a house at 25.0 m/s, perfectly
parallel to the ground below. How far has it traveled towards the
ground after 2.00 seconds? How far has it traveled away from
you (horizontally) after 2.00 seconds?
32. Practice Problem
•Yarrrr! An angry pirate launches a
cannonball at a ship over yonder. The
cannon is initially set at an angle of
25.0° and, with enough gunpowder,
can launch the cannonball at an initial
speed of 42.0 m/s along that angle.
The deck of the ship (with the cannon
on it) is 8.00 meters above the water.
•How much time will pass before the
cannonball lands in the water below?
•How far away from the ship will the
cannonball be when it lands?
