1. Vectors and Direction
In drawing a vector as an arrow you must choose a scale.
If you walk five meters east, your displacement can be represented by a 5 cm arrow pointing to the east.

2. Vectors and Direction
Suppose you walk 5 meters east, turn, go 8 meters north, then turn and go 3 meters west.
Your position is now 8 meters north and 2 meters east of where you started.
The diagonal vector that connects the starting position with the final position is called the resultant.

3. Vectors and Direction
The resultant is the sum of two or more vectors added together.
You could have walked a shorter distance by going 2 m east and 8 m north, and still ended up in the same place.
The resultant shows the most direct line between the starting position and the final position.

5. Calculate a resultant vector
1) You are asked for the resultant vector.
2) You are given three displacement vectors.
3) Make a sketch of the ant’s path, and add the displacement vectors by components.
4) Solve:
x1 = (-2, 0) m x2 = (0,3) m x3 = (6,0) m
x1 + x2 + x3 = ( ) m and ( ) m = (4,3) m
The final displacement is 4 meters east and 3 meters north from where the ant started.
An ant walks 2 meters West, 3 meters North, and 6 meters East.
What is the displacement of the ant?

6. 7.1 Finding Vector Components Graphically
Draw a displacement vector as an arrow of appropriate length at the specified angle.
Mark the angle and use a ruler to draw the arrow.

8. 7.1 Finding the Magnitude of a Vector
When you know the x- and y- components of a vector, and the vectors form a right triangle, you can find the magnitude using the Pythagorean theorem.

9. 7.1 Adding Vectors
Writing vectors in components make it easy to add them.

10. 7.1 Subtracting Vectors

11. 7.1 Calculate vector magnitude
A mail-delivery robot needs to get from where it is to the mail bin on the map.
Find a sequence of two displacement vectors that will allow the robot to avoid hitting the desk in the middle.
1) You are asked to find two displacement vectors.
2) You are given the starting and final positions.
3) The resultant vector must go from the start to the final position.
4) Solve:
The robot starts at (1,1) m and the mail bin is at (5,5) m.
The displacement required is (5,5) m - (1,1) m = (4,4) m.
First go up 4 meters, then over 4 meters.
x1 = (0,4) m, x2 = (4,0) m
Check the resultant: (4,0) m + (0,4) m = (4,4) m

12. 7.2 Projectile Motion and the Velocity Vector
Any object that is moving through the air affected only by gravity is called a projectile.
The path a projectile follows is called its trajectory.

13. 7.2 Projectile Motion and the Velocity Vector
The trajectory of a thrown basketball follows a special type of arch-shaped curve called a parabola.
The distance a projectile travels horizontally is called its range.

15. 7.2 Projectile Motion and the Velocity Vector
The velocity vector (v) is a way to precisely describe the speed and direction of motion.
There are two ways to represent velocity.
Both tell how fast and in what direction the ball travels.

16. 7.2 Calculate magnitude
Draw the velocity vector v = (5, 5) m/sec and calculate the magnitude of the velocity (the speed), using the Pythagorean theorem.
1) You are asked to sketch a velocity vector and calculate its magnitude (speed).
2) You are given the x-y component form of the velocity.
3) Set a scale of 1 cm = 1 m/sec to draw the sketch.
The Pythagorean theorem says a2 + b2 = c2.
4) Solve:
v2 = (5 m/sec)2 + (5 m/sec)2 = 50 m2/sec2
Take the square root and v = 7.07 m/sec

17. 7.2 Components of the Velocity Vector
Suppose a car is driving 20 meters per second.
The direction of the vector is 127 degrees.
The polar representation of the velocity is v = (20 m/sec, 127°).

18. 7.2 Calculate velocity
1) You are asked to calculate the components of the velocity vector.
2) You are given the initial speed and angle.
3) Draw a diagram and use vx = v cosθ and vy = v sinθ.
4) Solve: vx = (10 m/s)(cos 30o) = (10 m/s)(0.87) = 8.7 m/s
vy = (10 m/s)(sin 30o) = (10 m/s)(0.5) = 5 m/s
A soccer ball is kicked at a speed of 10 m/s and an angle of 30 degrees.
Find the horizontal and vertical components of the ball’s initial velocity.

19. 7.2 Adding Velocity Components
Sometimes the total velocity of an object is a combination of velocities.
One example is the motion of a boat on a river.
The boat moves with a certain velocity relative to the water.
The water is also moving with another velocity relative to the land.

20. 7.2 Adding Velocity Components

21. 7.2 Calculate velocity components
1) You are asked to calculate the resultant velocity vector.
2) You are given the plane’s velocity and the wind velocity.
3) Draw diagrams and add the components to get the resultant velocity.
4) Figure 7.13 shows the plane velocity vector.
vx = 100 cos 30o = 86.6 m/sec, vy = 100 sin 30o = 50 m/sec
Figure 7.14 shows the wind velocity vector.
vx = 40 cos 45o = 28.3 m/sec
vy = -40 sin 45o = m/sec
The resultant Z = ( , ) = (114.9, 21.7) m/sec.
An airplane is moving at a velocity of 100 m/s in a direction 30 degrees NE relative to the air.
The wind is blowing 40 m/s in a direction 45 degrees SE relative to the ground.
Find the resultant velocity of the airplane relative to the ground.

22. 7.2 Projectile Motion Vx Vy x y
When we drop a ball from a height we know that its speed increases as it falls.
The increase in speed is due to the acceleration gravity, g = 9.8 m/sec2.

23. 7.2 Horizontal Speed
The ball’s horizontal velocity remains constant while it falls because gravity does not exert any horizontal force.
Since there is no force, the horizontal acceleration is zero (ax = 0).
The ball will keep moving to the right at 5 m/sec.

24. 7.2 Horizontal Speed
The horizontal distance a projectile moves can be calculated according to the formula:

25. 7.2 Vertical Speed
The vertical speed (vy) of the ball will increase by 9.8 m/sec after each second.
After one second has passed, vy of the ball will be 9.8 m/sec.
After the 2nd second has passed, vy will be 19.6 m/sec and so on.

27. 7.2 Calculate using projectile motion
A stunt driver steers a car off a cliff at a speed of 20 meters per second.
He lands in the lake below two seconds later.
Find the height of the cliff and the horizontal distance the car travels.
1) You are asked for the vertical and horizontal distances.
2) You know the initial speed and the time.
3) Relationships that apply:
y = voyt - 1/2 gt2, x = v0xt
4) The car goes off the cliff horizontally, so voy = 0
y = - (1/2)(9.8m/s2)(2 s)2 = meters.
The negative sign shows the car is below its starting height.
Use x = voxt, to find the horizontal distance.
x = (20 m/s)(2 s) = 40 meters

28. 7.2 Projectiles Launched at an Angle
A soccer ball kicked off the ground is also a projectile, but it starts with an initial velocity that has both vertical and horizontal components.
*The launch angle determines how the initial velocity divides between vertical (y) and horizontal (x) directions.

29. 7.2 Steep Angle
A ball launched at a steep angle will have a large vertical velocity component and a small horizontal velocity.

30. 7.2 Shallow Angle
A ball launched at a low angle will have a large horizontal velocity component and a small vertical one.

31. 7.2 Projectiles Launched at an Angle
The initial velocity components of an object launched at a velocity vo and angle θ are found by breaking the velocity into x and y components.

32. 7.2 Range of a Projectile
The range, or horizontal distance, traveled by a projectile depends on the launch speed and the launch angle.

33. 7.2 Range of a Projectile
The range of a projectile is calculated from the horizontal velocity and the time of flight.

34. 7.2 Range of a Projectile
A projectile travels farthest when launched at 45 degrees.

35. 7.2 Range of a Projectile
The vertical velocity is responsible for giving the projectile its "hang" time.

36. 7.2 "Hang Time" You can easily calculate your own hang time.
Run toward a doorway and jump as high as you can, touching the wall or door frame.
Have someone watch to see exactly how high you reach.
Measure this distance with a meter stick.
The vertical distance formula can be rearranged to solve for time:

37. 7.2 Projectile Motion and the Velocity Vector
Key Question:
Can you predict the landing spot of a projectile?
*Students read Section 7.2 BEFORE Investigation 7.2

38. Marble’s Path Vy
x = ? y Vx
t = ?

39. In order to solve “x” we must know “t”
Y = vot – ½ g t2
vot = 0 (zero)
Y = ½ g t2
In the horizontal plane there is no force accelerating the marble. (Trace the motion of the marble as it travels and show that the marble would move in a straight line with the initial velocity of the photogate.
In the vertical plane, the force accelerating the marble is gravity, there is no initial velocity. (Trace the motion of the marble as you hold it at the photogate and drop it.)
2y = g t2
t2 = 2y g
t = 2y g

40. 7.3 Forces in Two Dimensions
Force is also represented in x-y components.

41. 7.3 Force Vectors
If an object is in equilibrium, all of the forces acting on it are balanced and the net force is zero.
If the forces act in two dimensions, then all of the forces in the x-direction and y-direction balance separately.

42. 7.3 Equilibrium and Forces
It is much more difficult for a gymnast to hold his arms out at a 45-degree angle.
To see why, consider that each arm must still support 350 newtons vertically to balance the force of gravity.

43. 7.3 Forces in Two Dimensions
Use the y-component to find the total force in the gymnast’s left arm.

44. 7.3 Forces in Two Dimensions
The force in the right arm must also be 495 newtons because it also has a vertical component of 350 N.

45. 7.3 Forces in Two Dimensions
When the gymnast’s arms are at an angle, only part of the force from each arm is vertical.
The total force must be larger because the vertical component of force in each arm must still equal half his weight.

46. 7.3 Forces and Inclined Planes
An inclined plane is a straight surface, usually with a slope.
Consider a block sliding down a ramp.
There are three forces that act on the block:
gravity (weight).
friction
the reaction force acting on the block.

47. 7.3 Forces and Inclined Planes
When discussing forces, the word “normal” means “perpendicular to.”
The normal force acting on the block is the reaction force from the weight of the block pressing against the ramp.

48. 7.3 Forces and Inclined Planes
The normal force on the block is equal and opposite to the component of the block’s weight perpendicular to the ramp (Fy).

49. 7.3 Forces and Inclined Planes
The force parallel to the surface (Fx) is given by
Fx = mg sinθ.

51. 7.3 Acceleration on a Ramp a = F m
Newton’s second law can be used to calculate the acceleration once you know the components of all the forces on an incline.
According to the second law:
Force (kg . m/sec2)
Acceleration
(m/sec2)
a = F m
Mass (kg)

52. 7.3 Acceleration on a Ramp Fx = m g sin θ a = F m
Since the block can only accelerate along the ramp, the force that matters is the net force in the x direction, parallel to the ramp.
If we ignore friction, and substitute Newtons' 2nd Law, the net force is:
Fx = m g
sin θ
a = F m

53. 7.3 Acceleration on a Ramp Fx = mg sin θ - m mg cos θ
To account for friction, the horizontal component of acceleration is reduced by combining equations:
Fx = mg sin θ - m mg cos θ

54. 7.3 Acceleration on a Ramp
For a smooth surface, the coefficient of friction (μ) is usually in the range
The resulting equation for acceleration is:

55. 7.3 Calculate acceleration on a ramp
1) You are asked to find the acceleration.
2) You know the mass, friction, force, and angle.
3) The relationships that apply
are: a = F÷m, Fx = mgsinθ
4) Calculate the x component of the skier’s weight:
Fx =(50 kg)(9.8 m/sec2)× (sin 20o) = N
Calculate the force:
F = N - 30 N = N
Calculate the acceleration:
a = N ÷ 50 kg = 2.75 m/sec2
A skier with a mass of 50 kg is on a hill making an angle of 20 degrees.
The friction force is 30 N.
What is the skier’s acceleration?

56. 7.3 Vectors and Direction Key Question:
How do forces balance in two dimensions?
*Students read Section 7.3 BEFORE Investigation 7.3

57. Application: Robot Navigation