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Physics · 06 / 12

Motion: position, velocity, acceleration

High school · intermediate · Worked example and 4 exercises with solutions, at your pace.

Need a foundation?Units: numbers have dimensionsEquations: keep the balance level
UnderstandSee an exampleGuided exercise3 progressive challenges

The principle, simply

Position locates a system in a reference frame. Velocity describes position change; acceleration describes velocity change. An axis sets the positive direction. Negative acceleration is not always slowing: velocity’s sign also matters. Standard kinematic formulas here require constant acceleration.

x=x_0+v_0*t+(a*t^2)/2
THE PICTURE TO REMEMBERMotion: position, velocity, acceleration
  1. 01Frame and axis
  2. 02Position x
  3. 03Velocity v
  4. 04Acceleration a

WORKED EXAMPLE

From principle to reasoning

x₀ = 0, v₀ = 2 m/s, a = 1 m/s², t = 4 s give x = 16 m and v = 6 m/s.

Read motion through slopes

For x(t)=x₀+v₀t+at²/2, position slope gives v(t)=v₀+at and velocity slope gives a. A negative-to-positive velocity means reversal, not necessarily return to the starting position.

Solve a problem in this subject

  1. Draw the system

    Define the studied object, reference frame and positive axis. A force diagram shows external interactions acting on the system.

  2. Sort quantities

    Separate data and unknowns; convert units before substitution. 36 km/h = 10 m/s, so never insert 36 into a formula expecting SI speed.

  3. Write the model

    Choose a law matching assumptions: constant acceleration, thin lens, ohmic resistor. Isolate the unknown symbolically before rounding.

  4. Interpret

    Check dimensions and signs. Negative velocity gives direction on the chosen axis. Discuss scale and model limitations.

YOUR TURN

01 · Try before checking

With v₀ = 10 m/s and a = −2 m/s², when is velocity zero?

One hint

Set v₀ + at to zero.

Reveal and understand the solution

0 = 10 − 2t gives t = 5 s. Continuing acceleration then predicts negative velocity.

Compare method, conditions and answer; identify the exact part to revisit.

IMPROVE THROUGH PRACTICE

Three additional challenges with solutions

Start with the first challenge. Write your method before opening a hint; use the solution to find the first step that needs work.

Reflect after practising
02 · Recognise and understandLink to this exercise

A car travels 150 m in 10 s. Find average speed.

A hint to get started

Average does not imply constant instantaneous speed.

Step-by-step solution
  1. v = distance/time = 150/10.
  2. v = 15 m/s.
  3. This measurement alone does not reveal accelerations.

Which step did you miss? Retry without help, then explain why the method works.

03 · Apply the methodLink to this exercise

v₀ = 3 m/s and constant acceleration is 2 m/s². Find v after 4 s.

A hint to get started

Use v = v₀ + at.

Step-by-step solution
  1. The speed gain is 2 × 4 = 8 m/s.
  2. v = 3 + 8 = 11 m/s.
  3. The positive sign follows the chosen acceleration direction.

Which step did you miss? Retry without help, then explain why the method works.

04 · Go furtherLink to this exercise

A body starts at rest with a = 2 m/s² for 3 s. Find its distance.

A hint to get started

Final speed times time would overestimate distance.

Step-by-step solution
  1. x − x₀ = v₀t + at²/2.
  2. The initial-speed term is zero, giving 2 × 9/2 = 9 m.
  3. Average speed is 3 m/s and final speed is 6 m/s.

Which step did you miss? Retry without help, then explain why the method works.

The trap to avoid

Signed displacement is not total distance when direction changes.

Without notes, can you…

  • Explain the principle in your own words?
  • Rebuild the diagram and its connections?
  • Solve the exercise and justify the method?

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