Kinematics is the study of motion without worrying about what causes it. We describe how objects move using four key quantities: displacement (Δx), velocity (v), acceleration (a), and time (t).
Distance is the total path length traveled and is always positive. Displacement is the change in position from start to finish — it has both magnitude and direction. If you walk 3 m east then 3 m west, your distance is 6 m but your displacement is 0 m.
Speed is how fast you move (scalar). Velocity is speed with a direction (vector). Average velocity = Δx / Δt. Instantaneous velocity is the velocity at a single moment.
Acceleration tells us how quickly velocity changes: a = Δv / Δt. Positive acceleration in the direction of motion means speeding up; negative acceleration (deceleration) means slowing down. On Earth, gravity provides a constant downward acceleration of g ≈ 9.8 m/s².
An object in free fall has only gravity acting on it. Near Earth’s surface a = −g = −9.8 m/s² (taking upward as positive). A ball thrown straight up slows by 9.8 m/s every second, stops at the top, then speeds up on the way down at the same rate.
Position–time graphs: the slope gives velocity. Velocity–time graphs: the slope gives acceleration and the area under the curve gives displacement. The exam frequently tests graph interpretation.
1) List knowns and unknowns. 2) Pick the equation containing exactly one unknown. 3) Substitute and solve. 4) Check units and the sign of your answer.
💡Kinematics describes motion using displacement, velocity, acceleration, and time. Master the five kinematic equations for constant acceleration — they are the backbone of mechanics.
📋 Key Formulas
📝 Worked Example 1
Example 1: A car accelerates from rest at 3 m/s² for 8 seconds. How far does it travel?
Step 1: Knowns: v₀ = 0 m/s, a = 3 m/s², t = 8 s
Step 2: Equation: Δx = v₀t + ½at²
Step 3: Δx = 0 + ½(3)(64) = 96 m
Answer: 96 m
📝 Worked Example 2
Example 2: A ball is thrown up at 20 m/s. Maximum height? (g = 9.8 m/s²)
Step 1: At top: v = 0. v₀ = 20, a = −9.8
Step 2: v² = v₀² + 2aΔx → 0 = 400 − 19.6Δx
Step 3: Δx = 400/19.6 ≈ 20.4 m
Answer: 20.4 m
📝 Worked Example 3
Example 3: A train at 30 m/s brakes at 2 m/s². Time to stop?
Step 1: v = v₀ + at → 0 = 30 − 2t → t = 15 s
Answer: 15 seconds
🧠No time variable? Use v² = v₀² + 2aΔx.
🧠Define positive direction first. Sign errors are the top mistake.
🧠Graph skills: slope of x-t = v, slope of v-t = a, area under v-t = displacement.
⚠️Forgetting g is negative when upward is positive.
⚠️Confusing distance and displacement in round-trip problems.
⚠️Applying kinematic equations when acceleration is not constant.
🎯 Try This Yourself
Open and read all sections to complete this module