Motion Graphs & Kinematic Equations — Practice · Grade 12 Science

Chapter: Motion & Kinematics

What this practice covers

By the end of Motion Graphs & Kinematic Equations you should be able to:

10 questions, numeric, multiple choice and written answer, each with a worked explanation.

Vocabulary used: position-time graph, velocity-time graph, slope, area under the curve, kinematic equations.

Problem 1

A velocity-time graph shows an object moving at a constant 6 m/s from t = 0 to t = 5.0 s. What displacement occurred during that time?
Answer and explanation

30 m

Displacement is the area under the velocity-time graph: 6 × 5.0 = 30 m.

Problem 2

A car starts from rest and accelerates at 3.0 m/s² for 4.0 s. Find its final velocity.
Answer and explanation

12 m/s

v = v₀ + at = 0 + (3.0)(4.0) = 12 m/s.

Problem 3

For that same car (from rest, 3.0 m/s² for 4.0 s), how far does it travel?
Answer and explanation

24 m

Δx = v₀t + ½at² = 0 + ½(3.0)(4.0²) = ½(3.0)(16) = 24 m.

Problem 4

On a position-time graph, a horizontal (flat) line means the object is…
  1. moving at constant positive velocity
  2. speeding up
  3. at rest (not moving)
  4. moving backward
Answer and explanation

at rest (not moving)

A flat line has zero slope, meaning zero velocity — the object's position isn't changing, so it is at rest.

Problem 5

A motorcycle traveling at 20 m/s brakes at a constant −5.0 m/s². How far does it travel before coming to a stop?
Answer and explanation

40 m

Use v² = v₀² + 2aΔx with v = 0: 0 = 20² + 2(−5.0)Δx, so 0 = 400 − 10Δx, giving Δx = 40 m.

Problem 6

Describe how you would find (a) an object's velocity from a position-time graph and (b) its displacement from a velocity-time graph.
Answer and explanation

(a) Velocity is the slope of the position-time graph — pick two points on the line and compute rise over run (change in position divided by change in time). (b) Displacement is the area between the velocity-time line and the time axis; for simple shapes you find the area of the rectangle, triangle, or trapezoid the graph forms.

Problem 7

A cart starts from rest at a=2 m/s² for 5 s. Find distance in one-direction motion. Show the reasoning, not just the conclusion.
Hints

These equations require constant acceleration over the interval.

Answer and explanation

Δx=½×2×25=25 m.

Problem 8

What does negative area under a velocity–time graph represent?
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

Negative displacement in the chosen coordinate; it does not mean negative elapsed time or negative path length.

Problem 9

Select an equation by known quantities and its assumptionsSelect an equation by known quantities and its assumptionsKnown quantityValueUseful relationshipInitial speed2 m/sv²=v₀²+2aΔxAcceleration3 m/s²Time not requiredDisplacement4 mv²=4+24=28v = +√28 ≈ 5.29 m/s for the stated forward motionSelect the physically relevant sign from the motion.
Original calculation with constant acceleration and positive initial motion.
Interpret the supplied visual. For these values, why is the positive velocity root appropriate?
Answer and explanation

The object begins moving forward and has forward acceleration across a positive displacement; it has not reversed.

The constant-acceleration relationships connect position, time and signed velocity. An equation without time is useful when time is not supplied, but taking a square root can introduce two algebraic signs. Choose the physically possible sign from the motion history and direction convention. Check that acceleration is modeled as constant over the interval; using a memorized equation without its conditions can produce an apparently precise result that describes the wrong situation.

The object begins moving forward and has forward acceleration across a positive displacement; it has not reversed.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the physical or environmental relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 10

A car starts at 5 m/s and accelerates constantly at 2 m/s² for 3 s. Find final speed and displacement, then verify displacement from the velocity–time area.
Answer and explanation

v=5+2×3=11 m/s. Δx=5×3+½×2×9=24 m. The velocity–time trapezoid has area (5+11)/2×3=24 m, an independent representation of the same model.

v=5+2×3=11 m/s. Δx=5×3+½×2×9=24 m.

The velocity–time trapezoid has area (5+11)/2×3=24 m, an independent representation of the same model.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the physical or environmental mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

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