VCE Vault

Year 10 Science · Physical and Earth Sciences · 18m read

Motion — Kinematics

Ever wonder how to predict exactly where a moving object will be in the next second? Kinematics is the toolkit that lets you do it—using equations and graphs to describe how things move. You'll learn to connect speed, acceleration, and time, then use motion graphs to visualize what's actually happening. This is core VCE Physics material, so mastering it now gives you a serious head start.

Learning objective: Apply equations of motion and analyse motion graphs.

By the end of this lesson you can:

  • I use v = u + at and similar equations.
  • I interpret motion graphs.
  • I solve motion problems.

Kinematics is reading a car's dashboard. The speedometer shows speed, the odometer distance, and pressing the accelerator changes velocity — acceleration.

Equations of motion

KEY VARIABLES:
• u — initial velocity (m/s).
• v — final velocity (m/s).
• a — acceleration (m/s²).
• t — time (s).
• s — displacement (m).

EQUATIONS (constant acceleration):
1. v = u + at.
2. s = ut + (1/2)at².
3. v² = u² + 2as.
4. s = (u + v)/2 × t.

CHOOSING THE RIGHT EQUATION:
• List what you KNOW and what you NEED.
• Pick the equation that includes those variables.

MOTION GRAPHS:
• Position–time (x–t):
– Slope = velocity.
– Straight line = constant velocity.
• Velocity–time (v–t):
– Slope = acceleration.
– Area under curve = displacement.
• Acceleration–time: area = change in velocity.

FREE FALL:
• On Earth, g ≈ 9.8 m/s² (often 10 m/s² for simplicity).
• Downward acceleration on falling/thrown objects (ignoring air resistance).

SIGN CONVENTIONS:
• Choose positive direction.
• Stick to it through all variables.

PROJECTILE MOTION (preview):
• Horizontal: constant velocity.
• Vertical: under gravity.
• Treat components SEPARATELY.

VECTORS:
• Velocity, displacement, acceleration are VECTORS (have direction).
• Speed and distance are SCALARS.

Use the variables list

  • Write u, v, a, t, s: Label known and unknown.
  • Match equation: Pick one that fits.
  • Watch signs: Consistency.

Quick reference

| Quantity | Formula | Unit |
|---|---|---|
| Speed | d/t | m/s |
| Velocity | Δs/t (with direction) | m/s |
| Acceleration | (v − u)/t | m/s² |
| Distance-time graph | Gradient = speed | |
| Velocity-time graph | Gradient = acceleration; area = displacement | |

Worked example (Medium)

A car accelerates from rest at 3 m/s² for 8 seconds. Find final velocity and distance.

  1. Known. u = 0, a = 3 m/s², t = 8 s.
  2. Final v. v = u + at = 0 + 3 × 8 = 24 m/s.
  3. Distance. s = ut + (1/2)at² = 0 + 0.5 × 3 × 64 = 96 m.

Answer: Final velocity 24 m/s; distance 96 m.

Kinematic Equations

  • Average speed: v = \dfrac{s}{t}
  • Acceleration: a = \dfrac{v-u}{t}
  • SUVAT displacement: s = ut + \dfrac{1}{2}at^2
  • SUVAT velocity squared: v^2 = u^2 + 2as
  • Average velocity: v_{\text{avg}} = \dfrac{u+v}{2}

Worked example (Hard)

A car decelerates from 20 m/s to rest at 4 m/s². Find: (a) time to stop, (b) stopping distance.

  1. Variables. u = 20 m/s, v = 0 (stopped), a = −4 m/s² (deceleration = negative).
  2. (a) Time. v = u + at → 0 = 20 − 4t → t = 5 s.
  3. (b) Stopping distance. v² = u² + 2as → 0 = 400 − 8s → s = 50 m.
  4. Verify. s = ut + (1/2)at² = 20(5) + (1/2)(−4)(25) = 100 − 50 = 50 m ✓

Answer: Time to stop: 5 s. Stopping distance: 50 m.

Common mistakes

  • Using speed when velocity is needed (ignoring direction) Velocity and acceleration are vectors. Applying the wrong sign for deceleration gives incorrect answers. Fix: Define a positive direction first. Deceleration is negative acceleration (opposite to motion).
  • Not including units in final answers Kinematics answers without units lose marks even if the number is correct. Fix: Always include: m for distance, m/s for velocity, m/s² for acceleration, s for time.
  • Reading a v-t graph: confusing area under curve with gradient GRADIENT of v-t = acceleration. AREA under v-t = displacement. Students often swap these. Fix: Gradient → acceleration (change in v / time). Area → displacement (base × height = s × m/s = m).

Practice questions

  1. The slope (gradient) of a velocity-time graph represents:

    1. Displacement
    2. Acceleration
    3. Speed
    4. Distance

    (1 marks)

  2. A ball is dropped from 20 m height. Using g = 10 m/s², find: (a) time to hit the ground, (b) speed at impact.

    (2 marks)

  3. A train accelerates from rest to 30 m/s in 60 s, travels at constant speed for 120 s, then decelerates to rest in 40 s. Find the total distance.

    (3 marks)

  4. Why does stopping distance increase proportionally with the SQUARE of speed?

    (3 marks)

Going Deeper — Projectile Motion

PROJECTILE MOTION: object launched with horizontal and vertical velocity components.

Key insight: horizontal and vertical motions are INDEPENDENT.

Horizontal: constant velocity (no air resistance). $x = u_x \cdot t$
Vertical: constant acceleration $g$ downward. $y = u_y t - \frac{1}{2}gt^2$

Example: Ball kicked horizontally at 15 m/s from a 20 m cliff (g = 10 m/s²).
Fall time: 20 = (1/2)(10)t² → t = 2 s.
Horizontal range: x = 15 × 2 = 30 m.
Final speed: √(15² + 20²) = √(625) = 25 m/s.

Applications: AFL kicks, artillery trajectories, satellite orbits.

Summary

  • v = s/t (average speed); a = (v−u)/t (acceleration).
  • SUVAT: v = u + at; s = ut + (1/2)at²; v² = u² + 2as.
  • v-t graph: slope = acceleration; area under curve = displacement.
  • Deceleration = negative acceleration (opposite to direction of motion).
  • Projectile: horizontal (constant v) and vertical (constant g) are independent.

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