Displacement, Velocity, and Acceleration: Study Guide (Physics C: Mechanics 1.2)

How this differs from the Physics 1 version. Physics C: Mechanics and Physics 1 are separate courses that both have a Topic 1.2 with the same title. This guide is the calculus-based one. It defines velocity and acceleration as derivatives, uses integrals to go back from acceleration to position, and handles acceleration that changes with time. The algebra-based treatment is in the Physics 1 study guide; do not mix the two when you revise.

Object, axis and displacement

We treat the moving thing as a point object: its size, shape and internal structure are ignored. Then we fix a coordinate system: an origin, a positive direction and a unit. Write it down every time, for example “+x to the right, origin at the start”. Every sign below depends on it.

Position x(t) is a function of time. Displacement over an interval is the change in position:

Δx = x − x₀

In two or three dimensions the same idea uses the position vector: Δr = r − r₀. This guide stays in one dimension, where a sign gives the direction completely.

Distance travelled is the total path length. It is never negative and equals |Δx| only if the object never turns around.

Average values

Averages use only the initial and final states over an interval Δt:

Average velocity: v_avg = Δx / Δt

Average acceleration: a_avg = Δv_x / Δt

An object is accelerating whenever its velocity changes, in size, in direction, or both.

From averages to derivatives

Shrink the interval. As Δt → 0, the average approaches the value at an instant:

v_x = lim (Δt → 0) Δx/Δt = dx/dt

a_x = lim (Δt → 0) Δv_x/Δt = dv_x/dt = d²x/dt²

Here is the limit in action for the motion in Worked example 1, x(t) = 2t³ − 9t² + 12t (x in m, t in s), at t = 0.50 s, where x = 4.0 m:

Δt (s)	Δx = x(0.50 + Δt) − x(0.50) (m)	Δx/Δt (m/s)
0.1	0.392	3.92
0.01	0.044402	4.4402
0.001	0.004494002	4.494002
→ 0		4.5 (= dx/dt at 0.50 s)

The derivative dx/dt = 6t² − 18t + 12 gives exactly 4.5 m/s at t = 0.50 s. On a graph, the secant slope becomes the tangent slope.

From derivatives back to position: integration

If you know a_x(t), you can rebuild the motion. Integrating undoes differentiating, but you need initial conditions to fix the constants:

v_x(t) = v_x0 + ∫₀ᵗ a_x dt

x(t) = x₀ + ∫₀ᵗ v_x dt

Graphically, the integral is the signed area between the curve and the time axis. Area below the axis counts as negative. So:

• Displacement from t₁ to t₂ = ∫ v_x dt (signed area under the v_x–t graph).
• Distance travelled = ∫ |v_x| dt (all areas counted as positive).
• Change in velocity = ∫ a_x dt (signed area under the a_x–t graph).

Deriving the constant-acceleration equations

When a_x is constant, integrate twice from t = 0:

1. v_x(t) = v_x0 + ∫₀ᵗ a_x dt = v_x0 + a_x t
2. x(t) = x₀ + ∫₀ᵗ (v_x0 + a_x t) dt = x₀ + v_x0 t + ½a_x t²
3. For the third equation, use the chain rule: a_x = dv_x/dt = (dv_x/dx)(dx/dt) = v_x (dv_x/dx). So a_x dx = v_x dv_x. Integrate both sides from (x₀, v_x0) to (x, v_x): a_x(x − x₀) = ½(v_x² − v_x0²), which gives v_x² = v_x0² + 2a_x(x − x₀).

Each step pulled a_x outside an integral. That is only allowed because a_x is constant. So these three equations are a limiting case of the general method, valid only for constant acceleration (for example, free fall near Earth’s surface, with g = 9.8 m/s² and air resistance ignored).

Speeding up or slowing down

Compare signs of v_x and a_x: same sign → speeding up; opposite signs → slowing down. The sign of a_x on its own tells you only the direction of the acceleration.

Worked example 1: from x(t) to v(t) and a(t)

Question. Take +x to the right. An object moves with x(t) = (2.0 m/s³)t³ − (9.0 m/s²)t² + (12 m/s)t for 0 ≤ t ≤ 3.0 s. Find v_x(t) and a_x(t), when it is at rest, when it speeds up or slows down, and its displacement and distance travelled.

1. Differentiate: v_x = dx/dt = 6.0t² − 18t + 12 = 6.0(t − 1)(t − 2) m/s.
2. Differentiate again: a_x = dv_x/dt = 12t − 18 m/s². It is zero at t = 1.5 s.
3. At rest when v_x = 0: t = 1.0 s and t = 2.0 s. Positions: x(1) = 5.0 m, x(2) = 4.0 m.
4. Sign chart:

Interval (s)	v_x	a_x	Motion
0 to 1.0	+	−	slowing down
1.0 to 1.5	−	−	speeding up (moving left)
1.5 to 2.0	−	+	slowing down
2.0 to 3.0	+	+	speeding up

5. Displacement: Δx = x(3) − x(0) = 9.0 m − 0 = +9.0 m.
6. Distance: legs |5 − 0| + |4 − 5| + |9 − 4| = 5 + 1 + 5 = 11 m.

Check. The integrals of v_x over [0, 1], [1, 2] and [2, 3] are +5, −1 and +5 m, matching Figure 2. Average velocity is 9.0 m ÷ 3.0 s = 3.0 m/s, while average speed is 11 m ÷ 3.0 s ≈ 3.7 m/s.

Worked example 2: non-constant acceleration by integration

Question. Take +x forward along a test track, origin at the start. A sled starts from rest at x₀ = 0. Its acceleration is a_x(t) = (4.0 m/s²) − (1.0 m/s³)t for 0 ≤ t ≤ 6.0 s. Find v_x(t), x(t), the velocity and position at 6.0 s, and the greatest velocity.

1. Velocity: v_x = 0 + ∫₀ᵗ (4.0 − 1.0t) dt = 4.0t − 0.50t² (m/s, t in s).
2. Position: x = 0 + ∫₀ᵗ (4.0t − 0.50t²) dt = 2.0t² − t³/6 (m).
3. At t = 6.0 s: v_x = 24 − 18 = 6.0 m/s; x = 72 − 36 = 36 m.
4. Greatest velocity: dv_x/dt = a_x = 0 at t = 4.0 s, so v_max = 16 − 8.0 = 8.0 m/s. After 4.0 s, a_x < 0 while v_x > 0, so the sled slows down but keeps moving forward.

Check the units. The constant 1.0 m/s³ times t (s) gives m/s², as an acceleration must.

Why the constant-acceleration equations fail here. Using x = ½a_x t² with the starting value a_x = 4.0 m/s² gives 72 m, twice the true 36 m. Acceleration falls during the run, so that equation does not apply.

Common misconceptions

• Using constant-acceleration equations for a(t). Integrate instead (Worked example 2).
• Forgetting initial conditions. ∫a_x dt gives a change in velocity. Add v_x0 to get v_x.
• “Negative a_x means slowing down.” Compare the signs of v_x and a_x.
• Treating ∫v_x dt as distance. It is displacement. For distance, split the integral where v_x = 0 and add the sizes.
• “v = 0 means a = 0.” At t = 1.0 s in Worked example 1, v_x = 0 but a_x = −6.0 m/s².
• Reading a secant slope as instantaneous velocity. Take the limit, or the derivative.
• Dropping units on coefficients. In x = 2.0t³, the 2.0 has unit m/s³.

Where this leads

Topic 1.3 (Representing Motion) builds on these graph and equation links, and Topics 1.4 and 1.5 extend them to reference frames and to two and three dimensions. Later, when acceleration depends on velocity, such as a_x = −kv_x for a drag force, you will separate variables and integrate; that belongs to Topic 2.9, Resistive Forces. Try the practice questions now, then use the revision notes and the checklist. You can also return to the course roadmap.