Difficulty: Introductory to Intermediate | Prerequisites: Trigonometry (sin, cos, tan), basic algebra
Vectors and one-dimensional kinematics form the backbone of everything in Physics 114. Vectors let you describe quantities that have both magnitude and direction (displacement, velocity, force), and kinematics gives you the equations that describe how objects move under constant acceleration. Nearly every problem on the midterm, from projectile motion to forces, builds on the skills in this section. You need to be able to decompose vectors into components, add them, and read motion graphs fluently before anything else will make sense.
To add vectors, break each into x and y components using cosine and sine, sum the components, then use the Pythagorean theorem for the resultant magnitude. For 1D motion, the slope of a position-time graph is velocity, the slope of a velocity-time graph is acceleration, and the area under a v-t curve is displacement. The four kinematic equations let you solve any constant-acceleration problem if you know three of the five variables (x, v₀, v, a, t).
Displacement
The straight-line distance and direction from the starting point to the ending point. In simple terms, it is how far you end up from where you started, regardless of the path you took.
Resultant vector
The single vector that has the same effect as two or more vectors combined. Think of it as the shortcut: if you walked two legs of a trip, the resultant is the crow-flies distance and direction from start to finish.
Vector components
The projections of a vector onto the x and y axes. For a vector of magnitude r at angle θ from the +x axis: x-component = r cos θ, y-component = r sin θ.
Position-time graph (x-t graph)
A graph plotting an object's position on the vertical axis against time on the horizontal axis. The slope at any point gives the instantaneous velocity.
Velocity-time graph (v-t graph)
A graph plotting velocity against time. The slope gives acceleration, and the area between the curve and the time axis gives displacement.
Constant acceleration
Motion where the acceleration does not change with time. This is the condition under which the standard kinematic equations apply.
ccw (counterclockwise)
Angles measured counterclockwise from the positive x-axis, which is the standard mathematical convention used throughout this course.
Break each vector into x and y components:
Aₓ = A cos θ_A, Aᵧ = A sin θ_A
Bₓ = B cos θ_B, Bᵧ = B sin θ_B
Sum the components: Rₓ = Aₓ + Bₓ, Rᵧ = Aᵧ + Bᵧ
Find the magnitude: R = √(Rₓ² + Rᵧ²)
Worked example (Exam Q3): A soccer player runs 10.0 m at 50.0° ccw, then jogs 13.0 m at 155.0° ccw.
Leg 1: x₁ = 10.0 cos 50° = 6.43 m, y₁ = 10.0 sin 50° = 7.66 m
Leg 2: x₂ = 13.0 cos 155° = −11.78 m, y₂ = 13.0 sin 155° = 5.49 m
Resultant: Rₓ = 6.43 + (−11.78) = −5.35 m, Rᵧ = 7.66 + 5.49 = 13.15 m
Magnitude: R = √(5.35² + 13.15²) = √(28.6 + 172.9) = √201.5 ≈ 14.2 m. Answer: D.
Initial positive velocity means the x-t graph starts with a positive slope (rising).
Constant negative acceleration means the slope decreases over time, and the curve bends downward (concave down).
Exam Q4: The correct graph is B, a concave-down curve that initially rises (positive velocity) but whose slope decreases as the car slows.
Graph A (concave up then levelling) suggests increasing then constant velocity. Graph C (concave down, starting from rest) suggests positive acceleration then slowing. Graph D (horizontal line) means zero velocity.
Displacement = area under the v-t curve (count areas below the axis as negative).
Worked example (Exam Q5): From the graph, between t = 0 and t = 15 s:
0 to 10 s: the velocity rises linearly from 0 to about 2 m/s. This is a triangle: area = ½ × 10 × 2 = 10 m.
10 to 15 s: the velocity rises from 2 m/s to 6 m/s. This is a trapezoid: area = ½ × (2 + 6) × 5 = 20 m.
Total displacement = 10 + 20 = 30 m. Answer: E.
The four kinematic equations connect five variables: initial velocity (v₀), final velocity (v), acceleration (a), displacement (Δx), and time (t).
Worked example (Exam Q6): A rocket starts from rest (v₀ = 0), reaches v = 26.0 m/s after Δx = 3.20 m.
Find acceleration: v² = v₀² + 2aΔx → 26.0² = 0 + 2a(3.20) → a = 676/6.40 = 105.6 m/s².
Find speed at t = 0.100 s: v = v₀ + at = 0 + 105.6 × 0.100 = 10.56 m/s ≈ 10.6 m/s. Answer: A.
R_x = \sum A_{ix} = \sum A_i \cos\theta_i \qquad R_y = \sum A_{iy} = \sum A_i \sin\theta_iR = \sqrt{R_x^2 + R_y^2} \qquad \theta_R = \tan^{-1}\!\left(\frac{R_y}{R_x}\right)v = v_0 + at\Delta x = v_0 t + \tfrac{1}{2}at^2v^2 = v_0^2 + 2a\,\Delta x\Delta x = \tfrac{1}{2}(v_0 + v)\,t\text{Displacement} = \text{area under } v\text{-}t \text{ curve}Vector addition is how navigators and pilots calculate a course when wind or current pushes them off a straight path. The kinematic equations describe any situation with constant acceleration, from a car braking on a highway to an elevator accelerating from rest.
Students often add vector magnitudes directly (10 + 13 = 23 m) instead of using components. This only works if both vectors point in exactly the same direction.
Confusing the slope of an x-t graph (velocity) with the slope of a v-t graph (acceleration). The slope tells you the rate of change of whatever is on the vertical axis.
Forgetting that area under the v-t curve can be negative if velocity is negative. Displacement is signed.
Choosing the wrong kinematic equation. Identify which of the five variables (v₀, v, a, Δx, t) you know and which you need, then pick the equation that contains exactly those.
⚠️ Vector addition by components is tested in at least one multiple-choice question every exam. Be fast with your calculator and watch the angle convention (ccw from +x).
⚠️ Expect a question showing several x-t or v-t graphs where you must match the graph to a verbal description of the motion. Practise reading concavity (concave up = accelerating positively, concave down = accelerating negatively).
⚠️ "Area under the curve" questions on v-t graphs are common. Break complex shapes into triangles, rectangles, and trapezoids.
True or false: The magnitude of the resultant of two vectors is always equal to the sum of their magnitudes. (False: only when they point in the same direction.)
Fill in the blank: The slope of a position-time graph gives the ____. (velocity)
True or false: A car with positive velocity and negative acceleration is slowing down. (True.)
Fill in the blank: The area under a velocity-time graph gives the ____. (displacement)
True or false: If an object starts from rest with constant acceleration, its speed after 0.2 s is twice its speed after 0.1 s. (True: v = at, so doubling t doubles v.)
Q: A hiker walks 5.0 km due east and then 8.0 km at 120° ccw from the +x axis. What is the magnitude of the net displacement?
A: x-components: 5.0 + 8.0 cos 120° = 5.0 + (−4.0) = 1.0 km. y-components: 0 + 8.0 sin 120° = 6.93 km. Magnitude = √(1.0² + 6.93²) = √(1 + 48.0) = 7.0 km.
Q: An object has a v-t graph that is a horizontal line at v = 4 m/s from t = 0 to t = 5 s, then drops linearly to 0 m/s at t = 10 s. What is the total displacement?
A: Rectangle area (0 to 5 s) = 4 × 5 = 20 m. Triangle area (5 to 10 s) = ½ × 5 × 4 = 10 m. Total = 30 m.
Q: A car accelerates uniformly from 12 m/s to 30 m/s over a distance of 210 m. What is the acceleration?
A: v² = v₀² + 2aΔx → 900 = 144 + 2a(210) → a = 756/420 = 1.8 m/s².
Q: Which kinematic equation does not contain time?
A: v² = v₀² + 2aΔx.
Vector addition is used directly in projectile motion (splitting initial velocity into horizontal and vertical components) and in force analysis (finding the net force from multiple forces at angles). The kinematic equations reappear in every free-fall and projectile problem. Understanding v-t graphs also prepares you for interpreting strobe diagrams and acceleration analysis in the forces section of the exam.
vector addition, component method, resultant vector, displacement, velocity, acceleration, position-time graph, x-t graph, velocity-time graph, v-t graph, area under the curve, kinematic equations, constant acceleration, SUVAT, ccw, counterclockwise, Physics 114, UW, midterm 1