Chapter 3Two-Dimensional Motion and Vectors

Chapter Objectives

• Distinguish Between a Scalar and a Vector• Add & Subtract Vectors• Determining Resultant Magnitude and

Direction• Apply Pythagorean Theorem and Tangent

Function to Vector Operations• Component Vectors• Recognize Examples of Projectile Motion• Apply Vectors to the Kinematic Equations

Scalar v Vector

• A scalar quantity is a physical quantity that has only a magnitude.

• Therefore a scalar quantity does not have a direction.

• Examples would be things that you count: time, people, speed, distance, etc.

• A vector quantity is a physical quantity that has a magnitude and direction.

• Examples would be a quantity that must show direction: displacement, velocity, acceleration, force, momentum, etc.

Drawing Vectors

• Vectors are drawn as arrows.• The arrow is called the head.• The other end is called the tail.• The head points in the direction of the stated vector quantity.• The tail is drawn where the vector starts.• You can start the vector anywhere, as long as you maintain a

consistent frame of reference.• Typically, the frame of reference is set up like a coordinate

plane with East being matched up with the positive x-axis.• Positive angle measures rotate in a counter-clockwise

fashion.• Negative angle measures rotate clockwise.

Examples

E

N

W

S

35o N of E

Notice that you state the line you rotated from last and the direction in which you rotated first.

40o N of W

Adding Vectors

• When adding vectors, place the vectors head-to-tail.• The sum of the vectors is the third leg of the triangle drawn

from the tail of the first vector to the head of the last vector.• Since this forms a triangle, we will need to use Trigonometry

rules to find the magnitude and direction.• The magnitude, or number, is found using Pythagorean

Theorem which is derived from the Law of Cosines.• The direction will be found using the Inverse Tangent.

• This will only work for right triangles.• Thanks to component vectors, we will be dealing with only right

triangles.

• Adding vectors follows the rules of standard mathematical addition, so it is commutative.

Example of Adding Vectors

E

N

W

S

35o N of E40o N of W

Let’s add the green with the purple.

Resultant vectors are drawn as dashed lines.

Inverse Tangent

Much like Pythagorean Theorem fits only with right triangles to find the magnitude of the resultant, the Inverse Tangent works only in right triangles to find direction.

tan Θ = opposite

adjacent

Θ

opposite adjacent

Solve for Θ by taking the inverse tangent of both sides.

Θ = tan-1opposite

adjacent

Component Vectors

Every vector is made of two component vectors. Component vectors are the vertical and horizontal portions of each vector.

v

vx

Θ

Notice that the component vectors are perpendicular, so we can use standard trig operations.

= v(cos Θ)

vyv(sin Θ) =

It is best to think of Θ as the angle made with the horizontal line. Otherwise, the parts of the triangle become different causing the sin and cos to switch.

Projectile Motion

• Projectile motion is defined as the motion of any object in two dimensions under the influence of gravity.

• Component vectors are necessary for use in calculating projectile motion.

• Projectiles follow the parabolic trajectories.• The kinematics equations can be used to

describe vertical and horizontal motion independently.

• Air resistance is ignored, so gravity becomes the only acceleration we need to account for.

Horizontal Components of Projectile Motion

• Since we ignore air resistance, we notice that objects travel the same displacement during each time interval.

• So we can assume that the horizontal acceleration is zero.• Thus the horizontal velocity is constant.

• So the one-dimensional kinematics equations are altered somewhat to fit horizontal two-dimensional motion.

v = v0 + a Δt Δx = 1/2(v + v0) Δt v2 = v02 + 2aΔx Δx = v0Δt + 1/2aΔt2

vx = v0,x Δx = vx Δt vx = v0,x Δx = vx Δt

Vertical Components of Projectile Motion

• The one-dimensional kinematics equations look very similar to the vertical two-dimensional versions.

• The only difference between the two is the vertical acceleration is…• Gravity

• g• So they look like

v = v0 + a Δt Δx = 1/2(v + v0) Δt v2 = v02 + 2aΔx Δx = v0Δt + 1/2aΔt2

vy = vy0 + gΔt Δy =1/2(vy + vy0)Δt vy2 = vy0

2 + 2gΔx Δy = vyoΔt + 1/2gΔt2

Projectiles Launched Horizontally

• Projectiles launched horizontally follow two conditions.• The initial velocity of the projectile is completely horizontal.

• So the distance from the edge of the cliff/table is found using a horizontal motion equation.

• The vertical motion is treated as if the object were dropped from rest.

• The time it takes to land is found using a vertical motion equation with the initial vertical velocity set equal to zero.

• The final velocity at any point during the path is found by summing the vertical and horizontal components of the velocity at that location.• This should be done by using Pythagorean Theorem.

Projectiles Launched at an Angle

• Whenever projectiles are launched at an angle, then we must treat the horizontal and vertical motions independent of one another.• This is because the acceleration is different

magnitude and direction for both.

• To accomplish this, we must break the initial velocity into vertical and horizontal components.

• Once that is accomplished, use the individual components to calculate the desired quantities using the kinematics equations.

Components and Kinematics

• Remember to find the horizontal component, use• vx = v cos Θ

• So that can change any formula you see that has vx in it.

• New notation will look like vx0

• That stands for initial velocity in the x-direction.

• Same goes for the vertical component.• vy = v sin Θ• vy0

• Initial velocity in the y-direction.

One-Dimensional v Two-Dimensional

One-DimensionalTwo-Dimensional

Horizontal Component

Two-Dimensional Vertical

Component

v = v0 + a Δtvx0 = vx

Velocity is constant horizontally.

vy = vy0 + gΔt

Δx = 1/2(v + v0) Δt Δx = vxΔtVelocity is constant

Δy = 1/2(vy + vy0) Δt

v2 = v02 + 2aΔx vx0 = vx

a=0, so that term disappearsvy

2 = vy02 + 2gΔy

Δx = v0Δt + 1/2aΔt2 Δx = vxΔta=0, so that term disappears

y = vy0Δt + 1/2gΔt2