1.8 Lesson:
Motion in 2D- An
Algebraic Approach
Addition of Perpendicular Vectors
• Two vectors acting perpendicular (900)
to each other may be added using
algebra instead of a scale diagram.
Resultant Displacement of Two
Perpendicular Vectors
1. Draw the vector diagram using the rules for
vector addition.
2. Use Pythagorean theorem to get the
magnitude of the resultant displacement.
3. Use tangent ratio to get the angle of the
direction of the resultant vector.
Example 1
A car drives 19.0km [S] and
then 9.0km [E]. What is the
resultant displacement of the
car?
∆𝑑 𝑇 = ∆𝑑1 + ∆𝑑2
∆𝑑 𝑇 = 19 𝑘𝑚 𝑆 + 9𝑘𝑚 [𝐸]
𝜃
∆𝑑1 = 19 𝑘𝑚[𝑆]
∆𝑑1
∆𝑑 𝑇
∆𝑑2
∆𝑑1
Applying Pythagoras theorem to
calculate the magnitude;
∆𝑑 2𝑇 = ∆𝑑12 + ∆𝑑22
∆𝑑 2𝑇 = 19 𝑘𝑚 2 + 9 𝑘𝑚 2
∆𝑑 2𝑇 = 361𝑘𝑚2 + 81 𝑘𝑚2
∆𝑑 𝑇 = 442 𝑘𝑚2
∆𝑑 𝑇 = 21.02 𝑘𝑚
Calculate the angle for the direction
∆𝑑2
𝑡𝑎𝑛 𝜃 =
∆𝑑2 = 9.0 𝑘𝑚[𝐸]
Resultant displacement is 21
km[S 250 E]
𝑡𝑎𝑛 𝜃 =
∆ 𝑑2
∆𝑑1
9 𝑘𝑚
= 0.474
19 𝑘𝑚
𝜃 = 𝑡𝑎𝑛−1 0.474 = 25.40
Components of a vector
• In the previous example you added two
perpendicular vectors to obtain one vector called the
resultant.
• The vector along the horizontal is called the xcomponent of the resultant vector.
• The vertical component of the resultant.
• A single vector can be broken into components or
components combined to form a single vector.
Directions
Example 1: What
are the
components of a
vector 30.0 m [E
250 N]
Set up the directions in
the same way as a yaxis versus x-axis
graph
• E is “+” x-component
• W is “-” x-component
• N is “+” y-component
• S is “-” y-component
Component Directions:
Example 1
• What are the
+𝑦[N]
components
of a vector
30.0 m [E 250
N]
−x[W]
sin 25° =
∆𝑑
∆𝑑𝑦
25°
∆𝑑𝑥
∆𝑑𝑦
cos 25° =
∆𝑑
∆𝑑𝑦 = ∆𝑑 𝑥 𝑠𝑖𝑛 25°
∆𝑑𝑦 = 30.0 𝑚 𝑥 0.423
∆𝒅𝒚 = 𝟏𝟐. 𝟕 𝒎 [N]
+𝑥[E]
∆𝑑𝑥
∆𝑑
∆𝑑𝑥 = ∆𝑑 𝑥 𝑐𝑜𝑠 25°
∆𝑑𝑥 = 30.0 𝑚 𝑥 0.906
−𝑦 [S]
∆𝒅𝒙 = 𝟐𝟕. 𝟐 𝒎 [E]
Components Method of Vector
Addition
1. Draw vectors from the origin of Cartesian
coordinate system
2. Break vector into components
2. Solve for total X-component.
3. Solve for total Y-component.
4. Use Pythagorean theorem to get the magnitude of
the resultant.
5. Use tangent ratio to get the angle of the direction.
Components With Angled
Directions:
If the x-direction is the reference then:
• To get the x-component use cosine
ratio
• To get the y-component use sine ratio
Example 2
A car drives 19.0km[S],
8.0km[E], 6.0km[N],
1.2km[W], and then 2.5km[N].
What is the resultant
displacement of the car?
A car drives 19.0km [W 30 ̊ N] and then goes 9.0km
[N 10 ̊ E] What is the resultant displacement of the car
using the components method?
+𝑦[N]
∆𝑑1𝑦
∆𝑑1𝑦 = ∆𝑑1 𝑠𝑖𝑛30°
= 19.0 𝑘𝑚 𝑥 0.5 = 9.5 km
∆𝑑2𝑥
∆𝑑2𝑥 = ∆𝑑2 𝑐𝑜𝑠80°
= 9.0 𝑘𝑚 𝑥 0.174 = 1.566 km
∆𝑑1
∆𝑑2
∆𝑑1𝑦
30°
−𝑥[W]
10° ∆𝑑
2𝑦
80°
∆𝑑1𝑥
∆𝑑2𝑥
−𝑦[S]
∆𝑑1𝑥 = ∆𝑑1 𝑐𝑜𝑠30°
∆𝑑1𝑥 = 19.0 𝑘𝑚 𝑥 0.866
∆𝑑1𝑥 = 16.345 km
+𝑥[E]
∆𝑑2𝑦 = ∆𝑑2 sin 80 °
∆𝑑2𝑦 = 9.0 𝑘𝑚 𝑥 0.985 =
∆𝑑2𝑦 = 8.865 km
A car drives 19.0km [W 30 ̊N] and then goes 9.0km
[N 10 ̊E] What is the resultant displacement of the car
using the components method?
+𝑦[N]
∆𝑑 𝑇
∆𝑑 𝑇𝑦 = ∆𝑑1𝑦 + ∆𝑑2𝑦
∆𝑑 𝑇𝑦 = 9.5 𝑘𝑚 + 8.865 𝑘𝑚
∆𝑑 𝑇𝑦 = 18.365 𝑘𝑚
∆𝑑 2𝑇 = ∆𝑑 2𝑇𝑥 + ∆𝑑 2𝑇𝑦
∆𝑑 2𝑇 = 14.779𝑘𝑚 2 + 18.365𝑘𝑚
∆𝑑 𝑇𝑦
18.365 km
∆𝑑 2𝑇 = 555.69 𝑘𝑚2
+𝑥[E]
−𝑥[W]
∆𝑑 𝑇𝑥
∆𝑑 𝑇 = 23.57 𝑘𝑚
14.779km
∆𝑑 𝑇𝑦 18.365 𝑘𝑚
−𝑦[S]
𝑡𝑎𝑛𝜃 =
=
∆𝑑 𝑇𝑥 14. 779 𝑘𝑚
∆𝑑 𝑇𝑥 = ∆𝑑1𝑥 + ∆𝑑2𝑥
tan 𝜃 = 1.48
∆𝑑 𝑇𝑥 = −16.345 𝑘𝑚 + 1.566 𝑘𝑚
𝜃 = 𝑡𝑎𝑛−1 1.24 = 51.2°
∆𝑑 𝑇𝑥 =− −14.779𝑘𝑚
Resultant displacement is 23.6
𝜃
km [W 510 N]
2
Example #4:
What is the resultant displacement
of an ATV that goes 2.0 km [N],
3.0 km [E], 3.5 km [SE], 5.0 km
[20 ̊ N of E], and finally 4.0 km
[30 ̊ E from N].
Adding Velocities-River Crossing
• A physics student has forgotten her lunch
and needs to return home to retrieve it. To do
so she hops into her motorboat and steers
straight across the river at a constant velocity
of 12 km/h [N]. If the river is 0.30 km
across and has no current, how long will it
take her to cross the river?
•
Relative Velocity
In the second diagram with no current the boat travels directly across the river in
a straight line. In the first diagram, the boat’s velocity directly across combined
with the river’s current, causes the boat to move diagonally across the river.
Adding Velocities-River Crossing
• A physics student has forgotten her lunch and needs to
•
•
•
•
return home to retrieve it. To do so she hops into her
motorboat and steers straight across the river at a constant
velocity of 12 km/h [N]. The river is 0.30 km across and
has current of 24 km/h [E].
Will she reach the shore at her home?
What is her resultant velocity across the river?
How long will it take her to cross the river?
How far from her home will she land?
Adding Velocities-River Crossing
• 𝑣𝑦 =
∆𝑑𝑦
• ∆𝑡 =
∆𝑑𝑦
• ∆𝑡 =
0.30 𝑘𝑚 [𝑁]
12 𝑘𝑚/ℎ[𝑁]
∆𝑡
𝑣𝑦
• ∆𝑡 = 0.025 ℎ 𝑜𝑟 1.5 𝑚𝑖𝑛
River Crossing-With Current
She will land east of her
home.
∆𝑑𝑥
River Crossing-With Current
𝑣𝑦 = 12 𝑘𝑚/ℎ(𝑟𝑖𝑣𝑒𝑟 𝑐𝑢𝑟𝑟𝑒𝑛𝑡)
𝑣𝑥 = 24 𝑘𝑚/ℎ(𝑏𝑜𝑎𝑡 ′ 𝑠 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦)
𝑣𝑇 =? (𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦)
∆𝑑𝑥
𝑣𝑇2 = 𝑣𝑦2 + 𝑣𝑥2
𝑣𝑇2 = 12𝑘𝑚/ℎ
2
+ 24𝑘𝑚/ℎ
𝑣𝑇2 = 720 𝑘𝑚/ℎ
θ
𝑣𝑇 =
720 𝑘𝑚/ℎ
𝑣 𝑇 =26.8 km/h
2
2
2
River Crossing-With Current
𝑣𝑥
𝑡𝑎𝑛𝜃 =
𝑣𝑦
∆𝑑𝑥
θ
24 𝑘𝑚/ℎ
𝑡𝑎𝑛𝜃 =
=2
12 𝑘𝑚/ℎ
𝜃 = 𝑡𝑎𝑛−1 2 = 63°
The resultant velocity of the boat is
27 km/h [N 630 E] (2significant
figures)
River Crossing-With Current
To calculate the time to cross the
river, focus on the component of
the velocity in the direction across
the river. This motion is uniform
velocity.
∆𝑑𝑥
𝑣𝑦 = 12 𝑘𝑚/ℎ[𝑁]
km [N]
∆𝑑
∆𝑑
∆𝑡 =
∆𝑡
𝑣
0.30 𝑘𝑚 [𝑁]
∆𝑡 =
= 0.025 ℎ
𝑘𝑚
12
[𝑁]
ℎ
Time to cross the river is 0.025 h
𝑣=
θ
∆𝑑𝑦 =0.30
River Crossing-With Current
The boat travels across the river
and along the river bank at the
same time. To calculate how far
from her home the student will
land, focus on the motion east.
𝑘𝑚
𝑣𝑥 = 24
∆𝑡 = 0.025 ℎ
ℎ𝐸
∆𝑑𝑥
θ
∆𝑑𝑥 =?
∆𝑑𝑥 = 𝑣𝑥 ∆𝑡
𝑘𝑚
∆𝑑𝑥 = 24
𝐸 𝑥 0.025 ℎ
ℎ
She will land 0.60 km east of her
home.