1. Kinematics in Two Dimensions
Chapter 3

2. Vectors vs. Scalars
Remember that vectors are quantities with both magnitude and direction, whereas scalars have only magnitude.
Examples: 65 mph north vs. 65 mph
velocity speed

3. Vector Addition
If the vector quantities are all the same direction, then adding them is done using simple math.
For example: A bicyclist travels 30 km south one day then another 35 km south the next day, his total displacement is 65 km south.

4. Vector Addition
If the vector quantities are on the same coordinate axis but opposite directions, add the opposite of one of the vectors to get the net or resultant vector.
Example: A bicyclist traveled 30 km south, then turned and traveled 35 km north.

5. Vector Addition
If we wanted to add vectors that are not on the same coordinate axis, we cannot use simple math.
Example: A bicyclist travels 25 km east and then travels 19 km north.

6. Vector Addition
Adding these vectors graphically is done with the use of a ruler and protractor.
Problem:
The lines drawn must be accurate.
This can be extremely difficult for the geometry-challenged population.

7. Vector Addition
The resultant displacement is represented by a new vector drawn from the point of origin to the ending point after the two vectors were arranged tip to tail.
The direction of the resultant displacement is determined by using a protractor.

8. Vector Addition
Another way to add vectors is to use Pythagorean theorem.
Take our same example: A bicyclist travels 25 km east and then travels 19 km north.
Draw a sketch. Notice how these two vectors form a right triangle.

9. Vector Addition DR2 = D12 + D22 Where: D1 = 25 km east
D2 = 19 km north DR D2  D1

10. Vector Addition
Notice how the sum of vectors not along the same line is smaller than their mathematical sum.
Remember that vectors MUST have both magnitude and direction.
Since we are working with a right triangle, the direction may be determined using the tan function, where tan  = opp/adj.

11. Vector Addition
Parallelogram method is also used…you’re really doing the same thing as adding the vectors tip-to-tail.

12. Adding Vectors by Components
Say we have a displacement vector with a magnitude of 45 km at 30o from the horizontal.
We can resolve this vector into its components that lie along the x and y axes.

13. Adding Vectors by Components
The vector is resolved into its x and y components by drawing dashed lines from the tip of the perpendicular to the axis lines.
The components are drawn along each axis where the dashed lines meet.

14. Adding Vectors by Components
We generally show vector components as arrows, but it is important to know that the components of a vector are scalar quantities.
We give the components + or – signs to indicate in what direction they are pointing.

15. Adding Vectors by Components
To add vectors using their components, we need to use trigonometric functions:
sine (sin)
cosine (cos)
tangent (tan)

16. Adding Vectors by Components
sin  = opposite side/hypotenuse
cosine  = adjacent side/hypotenuse
tangent  = opposite side/adjacent side

17. Adding Vectors by Components
Components of a vector are the projections of a vector along the axes of a coordinate system.
For example, let’s say that we had an airplane taking off at an angle of 20o from the horizontal with a velocity of 95 km/h, and we wanted to drive a dune buggy on the ground directly under the plane. How fast must the dune buggy go?

18. Adding Vectors by Components
Start solving this problem by writing down the given information, then draw a sketch…
Then analyze…what do we need to find?
Resolve the velocity vector so that we find the magnitude of its horizontal component. This will be the speed of the dune buggy on the ground.

19. Adding Vectors by Components
To add vectors that are not perpendicular to each other,
You must resolve each vector into its x and y components.
Get a total x and total y
Use Pythagorean theorem to find the resultant.

20. Projectile Motion
A practical application of using vectors and their components is projectile motion.
In order to analyze the motion of a projectile, we must look at it from the horizontal and vertical perspectives separately.

21. Projectile Motion
Projectile motion is described as the curved path that an object follows when thrown, launched, or otherwise projected near the surface of Earth.
The path of projectile is a parabola.
Projectiles do not just go straight up or down.

22. Projectile Motion
To keep things simple, we will ignore air resistance.
Because of this, we will assume that the horizontal velocity of a projectile is constant throughout its flight.
So, vox = constant, and ax = 0 m/s2

23. Projectile Motion
This also helps us to assume the vertical acceleration to be constant,
ay = g = m/s2.
We can apply the kinematic equations to an object being projected, but we must look at the x and y motion separately.

24. Projectile Motion
Galileo observed that an object projected horizontally will reach the ground in the same time as an object dropped vertically.
Look at figure 3-19 on p. 59

25. Projectile Motion
Projectiles are said to be freely falling bodies with a horizontal velocity.
So, they go up and down on the vertical axis much like the examples we studied last chapter.

26. Projectile Motion Check out the kinematic equations on p 60.
When we solve problems, it will help tremendously to draw a sketch.
Divide the given data into x and y motion.

27. Projectile Motion Example problem:
The Royal Gorge Bridge in Colorado rises 321 m above the Arkansas River. Suppose you kick a rock horizontally off the bridge. The magnitude of the rock’s horizontal displacement is 45.0 m. Find the speed at which the rock was kicked.

28. Projectile Motion Given: y = -321 m x = 45.0 m ay = -9.80 m/s2
Need: vox
Draw a sketch….

29. Projectile Motion
Because air resistance can be neglected, horizontal velocity is constant.
x = voxt

30. Projectile Motion
Because there is no initial vertical velocity, the following equation applies:
y = ½ ayt2

31. Projectile Motion
Remember, the time interval is the same for both the horizontal and vertical displacements, so the second equation can be rearranged to solve for t.
Next, rearrange the first equation to solve for vox.
Substitute the values and solve.