Vectors and Two-Dimensional Kinematics (Ch 3 & 4)
3.1 Components of Motion
Scalars vs. vectors:
Scalar: a quantity with only magnitude and units (e.g., distance, speed).
Vector: a quantity with magnitude and direction (e.g., displacement, velocity).
Intuition: to reach a location (like the library) you need both how far and which way.
The Components of a Vector:
Any vector can be resolved into perpendicular components in a 2D coordinate system (x and y).
This allows describing motion and forces along each axis separately.
3.1 Components of Motion: Position and velocity on a plane:
Position is described by two numbers (x, y).
Velocity can be described by components along x- and y-axes (vx, vy).
Magnitude and direction of a velocity vector:
If vx and vy are the components, the magnitude is
v = \sqrt{vx^2 + vy^2}
The direction angle relative to the x-axis is
\theta = \tan^{-1}\left(\dfrac{vy}{vx}\right)
Example: with components (vx, vy) = (4, 3) m/s, then v = \sqrt{4^2 + 3^2} = 5\ \text{m/s} and \theta = \tan^{-1}(3/4) \approx 36.87^{\circ}.
The components of displacement are obtained separately along x and y:
\Delta x = v_x \Delta t
\Delta y = v_y \Delta t
Equations of motion for two-dimensional kinematics (constant components):
x(t) = x0 + v{0x} t + \tfrac{1}{2} a_x t^2
y(t) = y0 + v{0y} t + \tfrac{1}{2} a_y t^2
Velocities along each axis:
vx(t) = v{0x} + a_x t
vy(t) = v{0y} + a_y t
Practice example (2D motion):
Initial velocity along x: v_{0x} = 1.50\ \mathrm{m/s}
Acceleration in y: a_y = 2.80\ \mathrm{m/s^2}
Starting at t_0 = 0\mathrm{s}, compute at t = 3.00\mathrm{s}:
Position: x = x0 + v{0x} t = x0 + 1.50(3) = x0 + 4.50\ \mathrm{m}
Position in y: y = y_0 + 0\cdot t + \tfrac{1}{2}(2.80)(3)^2 = 0 + 1.40 \cdot 9 = 12.6\ \mathrm{m}
Velocity components:
vx = v{0x} + ax t = 1.50 + ax t (if ax is given; in the example ax is not specified)
vy = v{0y} + a_y t = 0 + 2.80(3) = 8.40\ \mathrm{m/s}
Resultant speed: v = \sqrt{vx^2 + vy^2} (depends on a_x value)
Direction: \theta = \tan^{-1}\left(\dfrac{vy}{vx}\right)
3.2 Vector Addition and Subtraction
Vectors retain both magnitude and direction; addition/subtraction differ from scalars.
Two main methods:
Geometric/Graphical (Triangle Method)
Analytical Component Method
How to combine vector components to describe total motion:
Add components along each axis separately to obtain the resultant components, then construct the resultant vector.
Geometric method of vector addition (Triangle Method)
Steps:
Draw first vector A from the origin.
Draw second vector B from the tip of A.
Draw the vector from the tail of A to the tip of B. This is the resultant R.
This is a geometric way to visualize vector addition.
Note on the negative vector:
The negative of a vector has the same magnitude but opposite direction.
A + (−A) = 0.
Analytical Component Method
Steps:
Resolve all vectors into x- and y-components.
Add the x-components together: Fx = \sum F{x,i}
Add the y-components together: Fy = \sum F{y,i}
Use the Pythagorean theorem to obtain the final magnitude: |\mathbf{F}| = \sqrt{Fx^2 + Fy^2}
Determine the direction (angle) of the final vector using a trigonometric function, e.g.
\theta = \tan^{-1}\left(\dfrac{Fy}{Fx}\right)
Vector components and the analytical component method, example form:
If a vector A has components $(Ax, Ay)$ and B has components $(Bx, By)$, the resultant R has components:
Rx = Ax + B_x
Ry = Ay + B_y
Magnitude: |\mathbf{R}| = \sqrt{Rx^2 + Ry^2}
Direction: \theta = \tan^{-1}\left(\dfrac{Ry}{Rx}\right)
Unit-vector form:
Vectors can be written with unit vectors: \mathbf{F} = Fx \hat{i} + Fy \hat{j}
Example: if a vector a has components $(ax, ay)$, then \mathbf{a} = ax \hat{i} + ay \hat{j}
Projectile Motion
A projectile launched in an arbitrary direction has initial velocity components in both horizontal and vertical directions, but its acceleration is downward (gravity).
Acceleration is constant and vertical: \mathbf{a} = -g \hat{j} with g \approx 9.80\ \mathrm{m/s^2} (sign convention depends on axis).
Decomposition and independence:
Horizontal motion and vertical motion are analyzed separately.
Horizontal acceleration ax is typically zero (no air resistance), while vertical acceleration ay = -g.
Horizontal projection vs vertical drop (no air resistance):
Horizontal throw: constant v_x, vertical gravity affects y(t).
Both objects released from same height reach ground at the same time if only gravity acts and there is no air resistance, but their horizontal positions differ due to their horizontal velocities.
Projectile motion figures and clues:
Altitude and distance plots reflect the separation of horizontal and vertical motions.
Key results (no air resistance):
The launch angle that maximizes range on level ground is \theta = 45^{\circ}.
With air resistance the range is shortened and the maximum-range angle is less than 45^{\circ}.
Practice problem (example): An object launched from height with horizontal and vertical components:
Given: initial speed v_0 = 30.0\ \mathrm{m/s}, angle \theta = 35^{\circ}.
No air resistance: maximum height and range are:
Height: H = \dfrac{v_0^2 \sin^2 \theta}{2g}
Range: R = \dfrac{v_0^2 \sin 2\theta}{g}
Computed values (using g \approx 9.8\ \mathrm{m/s^2}):
H \approx \dfrac{(30)^2 \sin^2 35^{\circ}}{2\cdot 9.8} \approx 15.1\ \mathrm{m}
R \approx \dfrac{(30)^2 \sin(70^{\circ})}{9.8} \approx 86.4\ \mathrm{m}
Example: A stone thrown from a bridge into a river (downward angle) with given height and initial speed can be analyzed by resolving into components and solving for time to hit water, then comparing the horizontal distance traveled to the position of a block.
Relative Velocity
Velocity is relative to an inertial frame; measurements depend on the reference frame.
Example setup: velocities measured relative to ground (frame 1) vs. relative to a moving car (frame 2).
In two dimensions, velocity components and the angle relative to axes change with the reference frame.
Relative velocity and vector addition demonstration (conceptual):
If you know the velocity of one object in one frame and the velocity of the frame itself, you can find the velocity in another frame by vector subtraction/addition.
Practice problem (boat crossing a river):
Given: river width = 500 m, river flow rate (water velocity) = 2.55 km/h downstream, boat speed in still water = 8.00 km/h directly across.
Step 1: Determine velocity components relative to ground:
Across-the-river component (perpendicular to flow): v_x = 8.00\ \mathrm{km/h}.
Downstream component due to current: v_y = 2.55\ \mathrm{km/h}.
Step 2: Resultant velocity relative to stationary observer:
Magnitude: |\mathbf{v}| = \sqrt{vx^2 + vy^2} = \sqrt{8^2 + 2.55^2} \approx 8.41\ \mathrm{km/h}
Direction: \theta = \tan^{-1}\left(\dfrac{vy}{vx}\right) \approx \tan^{-1}(2.55/8) \approx 17.7^{\circ} downstream from across.
Step 3: Time to cross the river (width 500 m) given the across-the-river speed:
Convert across speed: 8.00\ \mathrm{km/h} = \dfrac{8000}{3600} \approx 2.22\ \mathrm{m/s}
Time to cross: t_{cross} = \dfrac{500}{2.22} \approx 225\ \mathrm{s}
Step 4: Downstream distance traveled during this time:
Convert downstream speed: 2.55\ \mathrm{km/h} = \dfrac{2550}{3600} \approx 0.708\ \mathrm{m/s}
Downstream distance: d{down} = vy t_{cross} \approx 0.708 \times 225 \approx 159\ \mathrm{m}
Therefore, velocity relative to stationary observer is about 8.41\ \mathrm{km/h} at an angle of about 17.7^{\circ} downstream from straight across, and the boat lands roughly 159\ \mathrm{m} downstream from the point directly opposite the start.
Review of Key Points (Chapters 3 & 4)
Two-dimensional motion is analyzed by considering each component separately; time is the common factor.
Vector components allow resolving motion into independent x- and y-directions, simplifying analysis of 2D kinematics.
In projectile motion, horizontal and vertical motions are treated separately; the overall path is a parabola due to the constant downward acceleration.
Range is the maximum horizontal distance traveled (no air resistance). The maximum occurs at a launch angle of \theta = 45^{\circ} (no air resistance).
With air resistance, the range is shortened and the optimal angle for maximum range is less than 45^{\circ}.
Relative velocity depends on the reference frame; velocity components and the resulting angle can change when viewed from a different inertial frame.
The use of unit vectors allows compact representation of vector quantities: \mathbf{F} = Fx \hat{i} + Fy \hat{j}.
The analytical component method provides a systematic way to add vectors by summing x- and y-components separately, then restoring magnitude and direction.
Practice problems illustrate how to apply these concepts to concrete scenarios, including projectiles from heights, cross-river motion, and vectors at angles.