1. AP Physics Chapter 3 Motion in Two Dimensions

2. 3.1 Components of Motion N E S W
Vectors are drawn as arrows on the coordinate plane.
- the length of the arrow corresponds to the magnitude
- the vector points to its direction A B C
1.5 m
3.0 m
60° β x y 0°
90°
180°
270°
Vector A has a magnitude of 3 meters and a direction of 60°above the positive x-axis.
Vector B has a magnitude of 3 meters and a direction of β° above the negative x-axis.
Vector C has a magnitude of 1.5 meters and a direction of 270°.

3. 3.1 Components of Motion
To analyze motion, a vector can be broken down into x- and y- components.
5 units
Ө = 53° y x
Given a vector v with a magnitude of a directed an angle Ө above the horizontal
vx = a cos Ө
vy = a sin Ө
y- component
x- component
example: Find the components of v if its magnitude is 5 units and Ө = 53°.
vx = a cos Ө = 5 cos 53° = 3 units
vy = a sin Ө = 5 sin 53° = 4 units
Note: The magnitude of the vector v is
v = √ vx2 + vy2 Why?

4. 3.1 Components of Motion
Kinematics problems in two-dimensions can be solved by:
resolving the displacement, velocity, and acceleration vectors into their respective components.
using the three constant acceleration equations separately for the x and y directions.
using the Pythagorean theorem to find the magnitude of your resultant vector.
using an inverse trig function to find the angle of your resultant vector

5. 3.1 Components of Motion
Example 1: A boat travels with a speed of 5.0 m/s in a straight path on a still lake. Suddenly, a steady wind pushes the boat perpendicularly to its straight line path with a speed of 3.0 m/s for 5.0 s. Relative to its position just when the wind started to blow, where is the boat at the end of this time?
Solution: List givens for x and y directions separately.
x-direction y-direction Unknowns: x and y
vo = 5.0 m/s vo = 3.0 m/s
a = 0 a = 0
Common to both: t = 5.0 s
Analysis: Both motions are motion with constant velocity. Choose the straight path of the boat as the x axis and the direction of the wind as the y axis.
Sketch:
wind y d
boat Ө x

6. Now, d = √ x2 + y2 = √ (25 m)2 + (15 m)2 = 29 m
x-direction y-direction Unknowns: x and y
vo = 5.0 m/s vo = 3.0 m/s
a = 0 a = 0
Common to both: t = 5.0 s
wind
boat y x d Ө
x-component: (use equation 1) x = vot + ½ at2 = (5.0 m/s)(5.0 s) + 0 = 25 m
y-component: (use equation 1) y = vot + ½ at2 = (3.0 m/s)(5.0 s) + 0 = 15 m
Now, d = √ x2 + y2 = √ (25 m)2 + (15 m)2 = 29 m
And Ө = tan-1 y = 15m = 31°
25 m

7. 3.2 Vector Addition and Subtraction
vector addition – combining vector quantities by using one of several techniques
• geometric methods (triangle, parallelogram, polygon)
• vector components and the analytical component method
vector subtraction – a special case of vector addition
• A - B = A + (-B)
• subtracting a vector is the same as adding a negative vector
resultant – the overall effect of combining vectors; the vector sum
Example: Add vectors A + B + C = R
Solution: Use tip-to-tail method

8. 3.2 Vector Addition and Subtraction
SCALE: 1 cm = 5 m
                                             
3.2 Vector Addition and Subtraction
Geometric (Graphical) Methods
use a convenient scale (ex: 1 cm = 10 meters)
To add vectors A and B, draw them tip-to-tail, starting at the origin.
The vector that extends from the tail of A to the tip of B, completing the triangle, is the resultant R = A + B.
When drawn to scale, the magnitude of R can be found by measuring R and using the scale conversion.
Use a protractor to measure the direction of the resultant, ӨR.
You may add on as many vectors as you like; the order doesn’t matter!
Example: Add three vectors graphically. Solution:

9. 3.2 Vector Addition and Subtraction
SCALE: 1 cm = 5 m
                                             
3.2 Vector Addition and Subtraction
Analytical (Trigonometry) Method
When vectors make a right triangle with each other,
use the Pythagorean theorem to find the magnitude of your resultant vector.
use the inverse trig function to find the angle of your resultant vector.
A = side adjacent to angle Ө
O = side opposite to angle Ө SOH – CAH - TOA
H = hypotenuse of triangle
sin Ө = O cos Ө = A tan Ө = O
A H A
Ө = sin-1 O Ө = cos-1 A Ө = tan-1 O
A H A H O Ө A

10. 3.2 Vector Addition and Subtraction
Example: Eric leaves the base camp and hikes 11 km, north and then hikes 11 km east. Determine Eric's resulting displacement and express in the magnitude-angle form.
Solution:
Magnitude of displacement = 16 km

11. 3.2 Vector Addition and Subtraction

12. 3.2 Vector Addition and Subtraction

13. 3.4 Projectile Motion
projectile – an object upon which the only force is gravity.
The key to projectile motion problems is recognizing the vertical and horizontal components of motion are independent.
• The vertical motion, or y- component is free fall, so a = m/s2
• The horizontal motion, or x- component, has zero acceleration. Why?
• Time is the common factor, and can be use to link the two equations.
Use kinematics equations 1 & 2 for the y- direction:
vy = vyo + a t
y = yo + vyo t + ½ ay t2
Use the same equations for the x- direction:
vx = vxo
x = vxo t

14. 3.4 Projectile Motion x- component of initial velocity vxo = vo cos Ө
y- component of initial velocity vyo = vo sin Ө

15. Given: horizontal motion vertical motion vxo = 120 m/s vyo = 0
Example: A package is dropped from an airplane travelling with a constant horizontal speed of 120 m/s at an altitude of 500 m. What is the horizontal distance the package travels before hitting the ground (range)?
Solution:
Given: horizontal motion vertical motion
vxo = 120 m/s vyo = 0
y = -500 m
Find: range
Since the range is given by x = vxo t , we have to find the time of flight t first.
From the vertical motion, we use y = vyot + ½ ayt2
So, -500 m = 0 + ½ (-9.80 m/s2) t2, solving, t = 10.1 s
Therefore, x = (120 m/s)(10.1 s) = 1.21 x 103 m = 1.21 km
Hint: The quantities such as initial velocities and displacements have to be treated independently. For example, the initial horizontal velocity is 120 m/s and the initial vertical velocity is zero. The 12.0 m/s can only be used in the horizontal motion and the 0 m/s can only be used in the vertical motion. A common mistake is to mix up these quantities or not treat them as independent. x
range
vo = vxo y
500 m

16. a) how long is the ball in the air? b) what is the range of the ball?
Example: A golfer hits a golf ball with a velocity of 35 m/s at an angle of 25° above the horizontal. If the point where the ball is hit and the point where the ball lands are at the same level,
a) how long is the ball in the air?
b) what is the range of the ball? x
range
vxo y
vyo vo Ө
Solution:
Given: horizontal motion vertical motion
vxo = vo cos Ө vyo = vo sin Ө
= (35 m/s) cos 25° = (35 m/s) sin 25°
= 31.7 m/s = 14.8 m/s
Find: a) t b) x
On landing, y = 0; and from y = vyot + ½ at2, we have
0 = (14.8 m/s)t + 1/2(-9.80 m/s2) t2. Solve using the quadric equation: t = 0 or 3.02 s.
The landing position is the second root, so the flight time is t = 3.0 s.
b) x = vxot = (31.7 m/s)(3.02 s) = 96 m
Alternate solution: use the range equation R = vo2 sin2Ө g

17. Check for Understanding:

18. vy = vyo + at ; vy = 0 at peak.
x = vxo t