Short on time? This page is the recap for the calculus-based course (separate from Physics 1, which uses no calculus). For explanations, graphs and worked examples, use the full study guide.
Recap
- Treat the object as a point. State the axis first, for example “+x to the right”.
- Displacement Δx = x − x₀ is signed. Distance travelled is the total path length.
- Averages use the initial and final states only: v_avg = Δx/Δt, a_avg = Δv_x/Δt.
- Instantaneous values are limits of averages as Δt → 0: derivatives.
- To go from a_x back to v_x and x, integrate and use the initial conditions v_x0 and x₀.
Key relationships
| Relationship | Units | Graph meaning |
|---|---|---|
| v_x = dx/dt | m/s | tangent slope of x–t graph |
| a_x = dv_x/dt = d²x/dt² | m/s² | tangent slope of v_x–t graph |
| v_x(t) = v_x0 + ∫₀ᵗ a_x dt | m/s | signed area under a_x–t graph gives Δv_x |
| x(t) = x₀ + ∫₀ᵗ v_x dt | m | signed area under v_x–t graph gives Δx |
| distance = ∫ |v_x| dt | m | all areas counted positive |
| v_x = v_x0 + a_xt; x = x₀ + v_x0t + ½a_xt²; v_x² = v_x0² + 2a_x(x − x₀) | — | constant a_x only |
Assumptions behind the numbers
- Point object, one straight line, inertial reference frame unless stated.
- The constant-acceleration equations assume a_x does not change. Free fall near Earth’s surface: a_y = −g with g = 9.8 m/s² (+y upward), air resistance ignored.
- Polynomial coefficients carry units (in x = 2.0t³, the 2.0 is in m/s³).
Mistakes to avoid
- Constant-a equations with a(t). Integrate instead.
- Missing v_x0 or x₀. An indefinite integral needs its constant.
- ∫v_x dt as distance. Split at v_x = 0 and add sizes.
- Negative a_x = slowing down. Compare signs of v_x and a_x.
- v_x = 0 so a_x = 0. Check a_x separately.
- Differentiating twice when asked for velocity.
- Using a velocity-dependent a(v) here. That needs separation of variables (Topic 2.9).
Quick self-check
- x(t) = 4.0t² − t³ (m). What is v_x at t = 2.0 s? (dx/dt = 8.0t − 3t² = 4.0 m/s)
- a_x = 2.0t (m/s²) and v_x0 = −3.0 m/s. What is v_x at t = 3.0 s? (−3.0 + 9.0 = 6.0 m/s)
- v_x = 2.0t − 4.0 (m/s). What are the displacement and distance from 0 to 4.0 s? (Displacement 0; distance 8.0 m)
Next: practice questions.