Notes on Motion, Vectors, and Projectile Motion
Motion and Reference Frames
All motion is relative; it depends on a reference frame.
A reference frame is the position from which an event is observed.
Relative speeds example (from transcript):
If an object moves at 200 km/h and the observer (reference) moves at 40 km/h in the same line, the observed relative speed can be 240 km/h (toward the target) or 160 km/h (away from the target), depending on direction.
Speed
Speed is the distance traveled during a specific unit of time.
Speed is a scalar quantity: described by magnitude only (no direction).
Constant speed: a fixed distance per unit time.
Average speed: includes the total distance and total time.
Key formulas:
S = \frac{d}{t}
S_{\text{avg}} = \frac{\text{total distance}}{\text{total time}}
Velocity
Velocity is the displacement of an object during a specific unit of time.
Displacement includes a direction; velocity therefore has direction as well.
Velocity is a vector (has magnitude and direction).
Speed with direction = velocity.
Key formulas:
\mathbf{v} = \frac{\Delta \mathbf{x}}{\Delta t}
Average velocity:
v{\text{avg}} = \frac{xf - xi}{tf - t_i}
Notes:
Velocity changes as position changes; displacement is a vector, while distance is a scalar.
Position vs. Time Graphs
The slope of a position-time graph represents velocity.
Position vs Time: vertical axis is position, horizontal axis is time.
The slope changes when velocity changes.
The steeper the slope, the greater the velocity.
The average velocity will generally differ from any instantaneous slope on the graph.
Any point on the line gives a corresponding instantaneous velocity.
Change in Direction
A change in direction on a position-time graph is shown by a switch from positive slope to negative slope, or from negative to positive.
A negative slope indicates motion toward the origin; a positive slope indicates motion away from the origin.
No motion is shown as a horizontal line (zero slope); position does not change.
Motion Maps
A motion map represents the position, velocity, and acceleration of an object at one-second intervals.
How to read:
Start with dots closest to the position line.
If the object moves away from the origin at a constant velocity, dots spread out at equal time intervals.
If it stops, dots stay in place for that interval.
If it then moves toward the origin at a slower constant velocity, spacing changes accordingly.
Lesson Goals
Differentiate between speed and velocity.
Describe motion of an object using different reference frames.
Solve speed and velocity problems using graphs and equations.
Interpret motion maps to describe linear motion.
Key Concepts
All motion is relative; it depends on a reference frame.
Speed is the distance traveled during a unit of time: S = \frac{d}{t}
Velocity is the displacement per unit time: \mathbf{v} = \frac{\Delta \mathbf{x}}{\Delta t}
Key Concepts (Position-Time Graphs)
A position-time graph shows the relationship of position over time for a moving object.
The slope represents the object's velocity: v = \frac{dx}{dt}
The slope changes when velocity changes; a steeper slope means higher speed.
Acceleration
Acceleration is the rate at which velocity changes over time.
Positive acceleration occurs when an object speeds up in the positive direction or slows down in the negative direction.
Negative acceleration occurs when an object slows down in the positive direction or speeds up in the negative direction.
When velocity changes, acceleration is present; when velocity is constant, acceleration is zero.
Acceleration Formulas (Constant Acceleration)
When acceleration is constant, several useful equations (SUVAT) can be derived:
General definitions:
a = \frac{\Delta v}{\Delta t}
v = v_0 + a t
x = x0 + v0 t + \tfrac{1}{2} a t^2
v^2 = v0^2 + 2a\,(x - x0)
Example: Final Position of a Train
Given: initial velocity v0 = 15\ \text{m/s}, acceleration a = 5\ \text{m/s}^2, time t = 10\ \text{s}, initial position x0 = 0\,\text{m}
Use: x = x0 + v0 t + \tfrac{1}{2} a t^2
Calculation:
x_f = 0 + (15)(10) + \tfrac{1}{2}(5)(10)^2 = 150 + 250 = 400\ \text{m}
Velocity vs. Time Graphs
A velocity-time graph shows how velocity changes over time.
The slope of a velocity-time graph represents acceleration:
a = \frac{\Delta v}{\Delta t} = \frac{dv}{dt}
The area under the velocity-time curve gives displacement:
\Delta x = \int v\,dt
For constant acceleration, a line with nonzero slope indicates changing velocity; a horizontal line indicates no acceleration (constant velocity).
The area under a velocity-time graph between t1 and t2 is the displacement over that interval, and for constant acceleration, \Delta x = \tfrac{1}{2}(v_0 + v) t where v is the final velocity after time t.
Displacement under Constant Acceleration
Total displacement during a motion with initial velocity v0 and acceleration a over time t can be found by several equivalent expressions:
\Delta x = v_0 t + \tfrac{1}{2} a t^2
\Delta x = \frac{v0 + vf}{2}\, t
More generally, for nonzero initial positions: x = x0 + v0 t + \tfrac{1}{2} a t^2
Note: For motion maps with changing acceleration or direction, different time intervals may show different slopes or changes in curvature.
Key Concepts (Acceleration)
Acceleration is the rate at which velocity changes over time: a = \frac{\Delta v}{\Delta t}
Positive acceleration: speed up in the positive direction or slow down in the negative direction.
Negative acceleration: slow down in the positive direction or speed up in the negative direction.
With constant acceleration, the above SUVAT equations apply.
The slope of a velocity-time graph corresponds to acceleration; the graph’s area relates to displacement.
Vectors: Basics
A vector is a quantity with both magnitude and direction.
Examples: displacement, velocity, acceleration.
Vectors are drawn with arrows; represented graphically or algebraically.
Two displacements are equal if they have the same magnitude and direction.
Two-Dimensional Vectors and Resultants
The resultant vector is the sum of two or more vectors.
If vectors are perpendicular, the magnitude of the resultant is given by:
R = \sqrt{A^2 + B^2}
If vectors form an angle (\theta), the magnitude is:
R = \sqrt{A^2 + B^2 + 2AB\cos\theta}
Components of Vectors
The components of a vector are the parts perpendicular to each other:
A_x = A\cos\theta
A_y = A\sin\theta
Resolution: given a vector, resolve into x and y components and sum components to get the resultant.
Sign of a component depends on the quadrant in which the vector lies:
First quadrant: both components positive
Second quadrant: Ax negative, Ay positive
Third quadrant: both negative
Fourth quadrant: Ax positive, Ay negative
Example: Components of Vectors
A truck travels 16.0 km on a straight road that is 40° north of east.
East is +x, North is +y.
Components:
A_x = 16.0\text{ km} \cos 40^{\circ} \approx 12.3\text{ km}
A_y = 16.0\text{ km} \sin 40^{\circ} \approx 10.3\text{ km}
Algebraic Addition of Vectors
If vectors are not perpendicular, resolve each into x and y components and add along each axis to get the resultant components:
Rx = Ax + Bx\,,\quad Ry = Ay + By
The resultant magnitude: R = \sqrt{Rx^2 + Ry^2}
The resultant direction: \tan\theta = \frac{Ry}{Rx}
Example (plane flying in two legs):
Leg 1: 182 km/h at 25° east of south (east component positive, south is negative y)
Leg 2: 170 km/h at 10° north of west (west is negative x, north is positive y)
Components (with x = east, y = north):
Leg 1: v{1x} = +182\sin 25^{\circ}, \quad v{1y} = -182\cos 25^{\circ}
Leg 2: v{2x} = -170\cos 10^{\circ}, \quad v{2y} = +170\sin 10^{\circ}
Over 1 hour for each leg, total components:
Rx = v{1x} + v_{2x} \approx 76.9 - 167.4 = -90.5\ \text{km/h}
Ry = v{1y} + v_{2y} \approx -165.0 + 29.5 = -135.5\ \text{km/h}
Resultant magnitude: R = \sqrt{(-90.5)^2 + (-135.5)^2} \approx 162.9\ \text{km/h}
Direction: angle from west toward south: \theta = \tan^{-1}\left(\frac{|Ry|}{|Rx|}\right) \approx \tan^{-1}\left(\frac{135.5}{90.5}\right) \approx 56^{\circ}
Therefore: approx. 56^{\circ}\text{ South of West}
Projectile Motion
Projectile motion is the curved motion from horizontal inertia combined with gravity pulling downward.
Key examples: ball off a table, cannonball, jump shot.
Projectiles follow a parabolic path; motion in horizontal and vertical directions is independent.
Horizontally Launched Projectiles
Horizontal motion:
Horizontal velocity is constant; acceleration in horizontal direction is zero.
Vertical motion:
Vertical velocity changes; acceleration is downward: a_y = -g\quad (g \approx 9.8\ \text{m/s}^2)
Horizontally Launched Projectiles (Equations)
Horizontal: vx = v{0x} = \text{constant}, x(t) = x0 + v{0x} t
Vertical: vy(t) = v{0y} - g t
Vertical position: y(t) = y0 + v{0y} t - \tfrac{1}{2} g t^2
Example: Pencil off a Desk (Horizontally Launched)
Given: height drop Ay = -0.76 m, horizontal distance Ax = 0.32 m, g = 9.8 m/s^2
Time of fall from height: t = \sqrt{\frac{2|Ay|}{g}} = \sqrt{\frac{2(0.76)}{9.8}} \approx 0.395\ \text{s}
Horizontal speed: v_{0x} = \frac{Ax}{t} = \frac{0.32\ \text{m}}{0.395\ \text{s}} \approx 0.81\ \text{m/s}
Projectiles Launched at an Angle
A launch with speed V0 at angle $\theta$ above the horizontal has initial components:
v{0x} = V0 \cos\theta
v{0y} = V0 \sin\theta
Horizontal motion:
vx = v{0x} = V_0 \cos\theta \quad (\text{constant})
x(t) = x0 + v{0x} t
Vertical motion:
vy(t) = v{0y} - g t = V_0 \sin\theta - g t
y(t) = y0 + v{0y} t - \tfrac{1}{2} g t^2 = y0 + V0 \sin\theta\, t - \tfrac{1}{2} g t^2
Key Concepts for Vectors and Projectile Motion
$V{ix} = V0 \cos\theta$, $V{iy} = V0 \sin\theta$ for launch at angle $\theta$.
Horizontal motion is independent of vertical motion for projectiles.
The magnitude of a resultant velocity in two dimensions uses vector components: $R = \sqrt{vx^2 + vy^2}$ at any instant.
Summary of Practical Steps for Vector Problems
Resolve vectors into x and y components:
Ax = A \cos\theta, \quad Ay = A \sin\theta
Sum components along each axis to get resultant components:
Rx = \sum Ax, \quad Ry = \sum Ay
Find magnitude and direction of the resultant:
R = \sqrt{Rx^2 + Ry^2}, \quad \theta = \tan^{-1}\left(\frac{Ry}{Rx}\right)
Quick Reference Formulas (Constant Acceleration)
Velocity with constant acceleration: v = v_0 + a t
Position with constant acceleration: x = x0 + v0 t + \tfrac{1}{2} a t^2
Velocity squared relation: v^2 = v0^2 + 2 a (x - x0)
Displacement under constant acceleration (alternative forms):
\Delta x = v_0 t + \tfrac{1}{2} a t^2
\Delta x = \frac{v_0 + v}{2}\, t
Notes on Reading the Content
The material emphasizes differentiating between speed (scalar) and velocity (vector).
It emphasizes interpreting motion via graphs (position-time and velocity-time) and via motion maps.
It introduces both one-dimensional motion and two-dimensional vector motion, including how to combine vectors and how to decompose them into components for addition.
It connects the kinematic equations to practical examples (train, pencil, plane, projectile motion).