Kinematics One-Dimensional Motion - Comprehensive Bullet-Point Notes
The Four Core Variables of Kinematics
Displacement (x or s): the change in position of an object.
Velocity (v): the rate of change of displacement.
Acceleration (a): the rate of change of velocity.
Time (t): the duration of motion.
Vector Placement Questions (quadrants and axes)
A Ԧ = −6 m i + 8 m j → x component negative, y component positive → Quadrant II (upper-left).
B = −3 m i − 4 m j → both components negative → Quadrant III (lower-left).
C Ԧ = 1.5 m/s i + ? j → x component positive; y component unknown from transcript; cannot determine quadrant without the y component value (data incomplete). If y component were zero, it would lie on the positive x-axis; if y component is positive, it would be in Quadrant I; if negative, Quadrant IV.
D = 6 m/s i − 8 m/s j → x positive, y negative → Quadrant IV (lower-right).
E ⃗ along i → Positive x-axis (x-axis only, no y component).
F Ԧ = −9.81 m/s² j → Negative y direction along the vertical axis (along the y-axis, downward if up is positive).
What is Kinematics?
Definition: the branch of mechanics that describes the motion of objects without considering the forces that cause the motion.
Focus in 1D: motion along a straight line, where all movement happens in a single direction or axis.
Key quantities: displacement, velocity, acceleration, and time.
The Four Core Variables (Detailed)
Displacement x or s: the change in position from an initial point to a final point.
Velocity v: the rate of change of displacement with respect to time.
Acceleration a: the rate of change of velocity with respect to time.
Time t: duration over which motion occurs.
Constant Velocity Formula
Formula: x = x_0 + v t
When to use: no acceleration (constant velocity).
Example: A motorboat travels at a steady 12.0 m/s along a river. How far will it go in 2.50 minutes?
Convert time: t = 2.50\text{ min} = 150\text{ s}.
Distance: \Delta x = v t = 12.0\times 150 = 1800\text{ m}.
Therefore, x = x_0 + 1800\text{ m}.
Definition of Acceleration
Formula: a = \frac{v - v_0}{t}
Meaning: change in velocity per unit time.
Example: A car accelerates from 15.0 m/s to 25.0 m/s in 4.0 s.
a = \frac{25.0 - 15.0}{4.0} = \frac{10}{4} = 2.5\ \text{m/s}^2.
First Equation of Motion: Velocity–Time Relation
Formula: v = v_0 + a t
Purpose: find final velocity after constant acceleration.
Example: A sprinter starts from rest (v_0 = 0) and accelerates at 4.2 m/s² for 3.50 s.
v = 0 + 4.2 \times 3.50 = 14.7\ \text{m/s}.
Second Equation of Motion: Position–Time Relation
Formula: x = x0 + v0 t + \tfrac{1}{2} a t^2
Purpose: find position after time t with acceleration.
Example: A skateboarder is moving at 2.0 m/s and accelerates uniformly at 1.5 m/s² for 6.0 s down a long driveway. How far does the skateboarder travel during this time?
Displacement: \Delta x = v_0 t + \tfrac{1}{2} a t^2 = (2.0)(6.0) + \tfrac{1}{2}(1.5)(6.0)^2 = 12.0 + 0.75\times 36 = 12.0 + 27.0 = 39.0\ \text{m}.
Therefore, x = x_0 + 39.0\ \text{m}.
Third Equation of Motion: Velocity–Displacement Relation
Formula: v^2 = v0^2 + 2 a (x - x0)
Purpose: relate velocity and displacement without time.
Example: A train is traveling at 25.0 m/s when it passes a 200 m mark along the track, it begins to slow down at 2.0 m/s² until it stops. Find the position where the train comes to rest.
At rest: v = 0, v0 = 25.0\ \text{m/s}, a = -2.0\ \text{m/s}^2, x0 = 200\ \text{m}.
Solve: 0 = (25.0)^2 + 2(-2.0)(x - 200)
0 = 625 - 4(x - 200)
x - 200 = \frac{625}{4} = 156.25\ \text{m}
x = 356.25\ \text{m}.
Formula Summary (x-axis)
The following formulas apply for motion along the x axis.
x axis formulas (what they include and when they apply):
x = x_0 + v t
Includes: x, x_0, v, t
Does not include: v_0, a (valid only when acceleration is zero, i.e., constant velocity)
x = x0 + v0 t + \tfrac{1}{2} a t^2
Includes: x, x0, v0, a, t
Does not include: v
v^2 = v0^2 + 2 a (x - x0)
Includes: v, v0, a, x, x0
Does not include: t
v = v_0 + a t
Includes: v, v_0, a, t
Does not include: x, x_0
Formula Summary (y-axis) with gravity
For vertical motion under gravity, using downward positive convention (g ≈ 9.81 m/s²):
y = y0 + v0 t - \tfrac{1}{2} g t^2
v^2 = v0^2 - 2 g (y - y0)
Includes: y, y0, v, t, g, v0 depending on the equation
Note: In these vertical equations, g is the acceleration due to gravity and appears with a sign depending on chosen up as positive or down as positive.
Example: Rock Dropped from Rest from a 45 m Cliff
Setup: rock released from rest, height h = 45 m, under gravity g = 9.81 m/s².
Use: s = \tfrac{1}{2} g t^2 with s = 45 m, solve for t.
Calculation: t = \sqrt{\dfrac{2 s}{g}} = \sqrt{\dfrac{2 \times 45}{9.81}} = \sqrt{\dfrac{90}{9.81}} \approx \sqrt{9.17} \approx 3.03\ \text{s}.
Result: time to hit ground ≈ 3.03 s.
Example: Ball Kicked Upward to Peak
Setup: ball launched upward with initial velocity v0 = 15 m/s, under gravity g = 9.81 m/s².
Use: At the peak, velocity v = 0; use v^2 = v_0^2 - 2 g h to find max height h.
Calculation: 0 = (15)^2 - 2 (9.81) h
Solve for h: h = \dfrac{(15)^2}{2 \times 9.81} = \dfrac{225}{19.62} \approx 11.46\ \text{m}.
Result: maximum height ≈ 11.5 m.
Assumptions and Limitations in Kinematics Problems
Acceleration is constant.
Motion is in a straight line.
Units are consistent (m, s, m/s, m/s²).
Air resistance is neglected.
Position-Time Graphs: Basics
Slope of the line = velocity.
Positive slope indicates forward motion; negative slope indicates backward motion.
Zero slope indicates stationary.
Curved line indicates changing velocity, i.e., acceleration.
Axes: Position x vs time t.
Position-Time Graphs: Examples
Straight line with positive slope → constant forward motion.
Straight line with zero slope → rest.
Curved upward → speeding up.
Curved downward → slowing down.
Velocity-Time Graphs: Basics
Slope of the line = acceleration.
Area under the curve = displacement.
Horizontal line above zero → constant positive velocity.
Horizontal line below zero → constant negative velocity.
Line sloping upward → speeding up.
Line sloping downward → slowing down.
Axes: Velocity v vs time t.
Velocity-Time Graphs: Examples
Constant acceleration: straight sloping line.
Constant velocity: flat line.
Changing acceleration: curve.
Example: Position–Time Graph for a Race Cart
Given a position–time graph for a race cart on a linear track.
Question: During which time interval did it first travel in a positive direction?
Answer: 0 – 10 s (positive slope initially).
Example: Position–Time Graph for a Race Cart – Negative Direction
Question: During which time interval did it first travel in a negative direction?
Answer: 15 – 40 s (slope becomes negative after 15 s).
Follow-up: What is the velocity at this moment?
Answer: v = -4 m/s at the moment when the direction first becomes negative (within 15 – 40 s interval).
Distance Traveled vs Displacement on Position–Time Graphs
Question: What is the total distance traveled?
Answer: 200 m (distance traveled accumulates absolute displacement).
Note: Displacement is the straight-line difference between final and initial positions; in this example it is 0 (start and end at the same position).
Speed–Time Graphs: 50 s Car Journey
Task: Describe the journey and determine the acceleration for each section from the slope (since slope = acceleration).
Given accelerations for sections:
A = 1.5 m/s²
B = 0 m/s²
C = 1.0 m/s²
D = −1.25 m/s²
Additional: Total distance traveled is found by the area under the speed–time graph.
Example: Speed–Time Graph Distances (Total Distance)
For the 50 s car journey, the calculated total distance traveled is 675 m, derived from summing the areas under the graph across all sections.
Segment summaries given:
A → Δx contributed by area under the velocity curve for section A
B → zero acceleration implies constant velocity for section B
C → additional area under velocity curve for section C
D → negative acceleration contributing to deceleration/velocity reduction
Final total distance: 675 m.
Quick Reference: Key Equations (All in LaTeX)
Constant velocity: x = x_0 + v t
Acceleration: a = \frac{v - v_0}{t}
Velocity with constant acceleration: v = v_0 + a t
Position with constant acceleration: x = x0 + v0 t + \tfrac{1}{2} a t^2
Velocity from displacement: v^2 = v0^2 + 2 a (x - x0)
Vertical motion under gravity (up is positive): y = y0 + v0 t - \tfrac{1}{2} g t^2
Velocity from vertical displacement: v^2 = v0^2 - 2 g (y - y0)
Solve time from height with gravity: t = \sqrt{\dfrac{2 s}{g}}
Peak height from upward launch: h = \dfrac{v_0^2}{2 g}
Notes on Real-World Relevance
These equations underpin everyday measurements of motion: driving, sports, machinery, and safety analyses.
Understanding when to use which equation avoids unnecessary complexity in solving problems.
Graphical interpretations (slope, area) provide intuitive checks for algebraic results.