Chapter 3

Two-Dimensional Motion and Vectors

Chapter Objectives

Distinguish Between a Scalar and a VectorAdd & Subtract VectorsDetermining Resultant Magnitude and DirectionApply Pythagorean Theorem and Tangent Function to Vector OperationsComponent VectorsRecognize Examples of Projectile MotionApply 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 is found using Pythagorean Theorem which is derived from the Law of Cosines.

The direction will be found using the Inverse Tangent or a more reliable form, Law of Sines.

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 TangentMuch 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.

vf = vi + a Δt Δx = 1/2(vf + vi) Δt vf2 = vi

2 + 2aΔx Δx = viΔt + 1/2aΔt2

vxf = vxi Δx = vxi Δt vxf = vxi Δx = vxi 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

gSo they look like

vf = vi + a Δt Δx = 1/2(vf + vi) Δt vf2 = vi

2 + 2aΔx Δx = viΔt + 1/2aΔt2

vyf = vyi + gΔt Δy =1/2(vyf + vyi)Δt vyf2 = vyi

2 + 2gΔy Δy = vyiΔt + 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 = vi cos Θ

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

New notation will look like vxi That stands for initial velocity in the x-direction.

Same goes for the vertical component. vy = vi sin Θ vyi

Initial velocity in the y-direction.

One-Dimensional v Two-Dimensional

One-Dimensional

Two-Dimensional Horizontal

Component

Two-Dimensional Vertical

Component

vf = vi + a Δt vxi = vxf

Velocity is constant horizontally.

vyf = vyi + gΔt

Δx = 1/2(vf + vi) Δt Δx = vxΔtVelocity is constant

Δy = 1/2(vyf + vyi) Δt

vf2 = vi

2 + 2aΔx vxi = vxf

a=0, so that term disappears

vyf2 = vyi

2 + 2gΔy

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

Δy = vyiΔt + 1/2gΔt2