Displacement, Velocity and Motion Graphs

Key Terms

Displacement x(t): : the position of a particle relative to an origin, at time t. Can be positive, negative or zero.
Velocity v(t) = x′(t) = dx⁄dt: : the instantaneous rate of change of displacement. Positive = moving right/up; negative = moving left/down.
Acceleration a(t) = v′(t) = x″(t) = d²x⁄dt²: : rate of change of velocity.
Particle at rest: : v(t) = 0.
Moving right (positive direction): : v(t) > 0.
Moving left (negative direction): : v(t) < 0.
Initial values: : x(0) = initial displacement; v(0) = initial velocity.
Distance ≠ Displacement: : distance is total path length (always ≥ 0); displacement is change in position (can be negative).

📚 Key Relationships
x(t) → differentiate → v(t) = x′(t)
v(t) → differentiate → a(t) = v′(t)
Particle at rest: v(t) = 0
Direction change: v changes sign

Hot Tip When v(t) changes sign, the particle changes direction. To find total distance (not displacement), split the motion at these sign changes and add path lengths separately.

Worked Example 1 — Velocity and Acceleration

Question: A particle moves with displacement x = t³ − 6t² + 9t. Find v(t), a(t), and the times when the particle is at rest.

v(t) = x′(t): v(t) = 3t² − 12t + 9 = 3(t² − 4t + 3) = 3(t − 1)(t − 3)

a(t) = v′(t): a(t) = 6t − 12

At rest (v = 0): 3(t − 1)(t − 3) = 0 → t = 1 or t = 3.

x(1) = 1 − 6 + 9 = 4 | x(3) = 27 − 54 + 27 = 0

Worked Example 2 — Distance vs Displacement

Question: Using x = t³ − 6t² + 9t above, find the total distance travelled from t = 0 to t = 4.

Particle reverses at t = 1 and t = 3. Positions: x(0) = 0, x(1) = 4, x(3) = 0, x(4) = 4.

Segment 0→1: moves from 0 to 4 → distance = 4

Segment 1→3: moves from 4 to 0 → distance = 4

Segment 3→4: moves from 0 to 4 → distance = 4

Total distance = 12 units. Displacement = x(4) − x(0) = 4.

Introduction: Motion Along a Line

Calculus gives us a powerful language for describing motion. When a particle moves along a straight line, we track its displacement x(t) — its signed distance from a fixed origin. Differentiating displacement gives velocity; differentiating velocity gives acceleration.

The Hierarchy of Motion

The three quantities are linked by differentiation:

x(t) ⟶ differentiate ⟶ v(t) = x′(t) ⟶ differentiate ⟶ a(t) = v′(t)

Sign Conventions

Choose a positive direction (usually right or up). A negative displacement means the particle is to the left of/below the origin.
v > 0: moving in the positive direction; v < 0: moving in the negative direction; v = 0: stationary.
a > 0: velocity is increasing; a < 0: velocity is decreasing (could still be moving in positive direction if v is large and positive).

Reading Motion Graphs

Displacement-time graph: gradient = velocity; horizontal tangent (zero gradient) = particle at rest.
Velocity-time graph: gradient = acceleration; sign of v = direction of motion; zero crossing = change of direction.

Worked Example — Initial Position and Velocity

x(t) = 2t³ − 9t² + 12t − 4. Find initial position, initial velocity, and times at rest.

Step 1: x(0) = −4. Initial position is 4 units to the left of the origin.

Step 2: v(t) = 6t² − 18t + 12. v(0) = 12 m/s (moving in positive direction initially).

Step 3: v(t) = 0: 6t² − 18t + 12 = 0 → t² − 3t + 2 = 0 → (t−1)(t−2) = 0 → t = 1 or t = 2.

Worked Example — Distance vs Displacement

For x(t) = t² − 4t, find the total distance from t = 0 to t = 4.

Step 1: v(t) = 2t − 4 = 0 at t = 2 (direction change).

Step 2: x(0) = 0, x(2) = 4 − 8 = −4, x(4) = 16 − 16 = 0.

Step 3: 0 → −4: distance = 4; −4 → 0: distance = 4.

Total distance = 8 units. Displacement = x(4) − x(0) = 0.

💡 Key Reminder: Total distance requires you to split the motion wherever v = 0 (direction change). Displacement simply = x(final) − x(initial).

Summary

In kinematics: velocity is the derivative of displacement, and acceleration is the derivative of velocity. The sign of v tells you direction; v = 0 means the particle is momentarily at rest (which may indicate a direction change). Distance ≠ displacement unless the particle never reverses.

Mastery Practice

See Answers ➔

Fluency
A particle has displacement x = t² − 4t + 3 (in metres, t in seconds).
1. (a) Find the velocity function v(t).
2. (b) Find the acceleration function a(t).
3. (c) Find the velocity and acceleration at t = 3 s.
Fluency
A particle moves with displacement x = t³ − 6t² + 9t (in cm, t in seconds).
1. (a) Find the initial displacement and initial velocity.
2. (b) Find the times when the particle is at rest.
3. (c) Find the displacement at each time the particle is at rest.
Fluency
For the displacement function x = t² − 2t − 3:
1. (a) Find v(t).
2. (b) Find when the particle is moving in the negative direction (v < 0).
3. (c) Find when the particle is moving in the positive direction (v > 0).
Understanding
A particle has displacement x = 2t³ − 9t² (in metres, t ≥ 0 seconds).
1. (a) Find the times when the particle is at rest.
2. (b) Determine whether the particle changes direction at each of these times.
3. (c) Find the displacement at each time the particle is at rest.
Understanding
A ball thrown upward has height h = 20t − 5t² metres (t in seconds).
1. (a) Find the maximum height reached by the ball.
2. (b) Find when the ball hits the ground (h = 0).
3. (c) What is the velocity when it hits the ground?
Understanding
A particle has velocity v = 6t² − 4t m/s.
1. (a) Find the acceleration function a(t).
2. (b) Find the acceleration at t = 2 s.
3. (c) Find when the particle is at rest.
Understanding
A particle moves with x = t² − 4t from t = 0 to t = 4 seconds.
1. (a) Find when the particle changes direction.
2. (b) Calculate the total distance travelled over this interval.
3. (c) Calculate the displacement over this interval.
4. (d) Explain why distance and displacement are different here.
Problem Solving
Two particles start from the origin at t = 0. Particle A has position x_A = t² + 2t and particle B has position x_B = 3t (both in metres, t in seconds).
1. (a) Find when the two particles are at the same position (after t = 0).
2. (b) Find the velocities of each particle at that time.
3. (c) Which particle is moving faster at that moment?
Problem Solving
A particle has displacement x = t³ − 3t² + 3t + 1.
1. (a) Find v(t) and simplify by completing the square.
2. (b) Show that v(t) ≥ 0 for all t, and hence the particle never reverses direction.
3. (c) Find a(t) and determine when acceleration is zero.
Problem Solving
Reading a displacement-time graph.

Described graph: A particle starts at x = 3 at t = 0, moves to x = 0 at t = 2 (passing through origin), continues to x = −1 at t = 3 (minimum), then returns to x = 3 at t = 6.
1. (a) At approximately what time(s) is the velocity zero? Explain how you can tell from the graph.
2. (b) Over which intervals is the velocity negative?
3. (c) At which moment is the speed (|v|) greatest? Explain.
4. (d) Estimate the total distance travelled from t = 0 to t = 6.

Displacement, Velocity and Motion Graphs

Key Terms

Worked Example 1 — Velocity and Acceleration

Worked Example 2 — Distance vs Displacement

Introduction: Motion Along a Line

The Hierarchy of Motion

Sign Conventions

Reading Motion Graphs

Summary

Mastery Practice

A particle has displacement x = t² − 4t + 3 (in metres, t in seconds).

A particle moves with displacement x = t³ − 6t² + 9t (in cm, t in seconds).

For the displacement function x = t² − 2t − 3:

A particle has displacement x = 2t³ − 9t² (in metres, t ≥ 0 seconds).

A ball thrown upward has height h = 20t − 5t² metres (t in seconds).

A particle has velocity v = 6t² − 4t m/s.

A particle moves with x = t² − 4t from t = 0 to t = 4 seconds.

Two particles start from the origin at t = 0. Particle A has position xA = t² + 2t and particle B has position xB = 3t (both in metres, t in seconds).

A particle has displacement x = t³ − 3t² + 3t + 1.

Reading a displacement-time graph.

Two particles start from the origin at t = 0. Particle A has position x_A = t² + 2t and particle B has position x_B = 3t (both in metres, t in seconds).