Motion & Kinematics — Problem set, Grade 12

Describe motion with graphs, vectors, and bounded constant-acceleration models.

Review the key ideas

Describing Motion

Position is measured from a chosen origin; distance is total path length (a scalar) while displacement is the straight-line change in position (a vector), and every measurement is made relative to a frame of reference.

Velocity & Acceleration

Acceleration is velocity change per time. Near Earth, ideal free fall has approximately constant downwardg; negative acceleration does not always mean decreasing speed.

Motion Graphs & Kinematic Equations

On a position-time graph slope = velocity; on a velocity-time graph slope = acceleration and area = displacement; and four kinematic equations (v = v₀ + at, Δx = v₀t + ½at², v² = v₀² + 2aΔx, Δx = ½(v₀ + v)t) solve any constant-acceleration problem.

Projectile & Two-Dimensional Motion

Uniform gravity with negligible drag gives separate component equations sharing time. The 45° range maximum also assumes fixed launch speed and equal launch/landing heights.

Problem 1

A hiker walks 5 km north, then 12 km east. Find the magnitude of her displacement.
Answer and explanation

13 km

The two legs are perpendicular, so use the Pythagorean theorem: displacement = √(5² + 12²) = √(25 + 144) = √169 = 13 km.

Problem 2

A runner sprints 100 m north, then jogs 100 m back south to her exact starting point. Find her total distance and her displacement.
Answer and explanation

distance 200 m; displacement 0 m

Distance adds the whole path: 100 + 100 = 200 m. Displacement is the straight-line change in position; since she returns to the start, it is 0 m.

Problem 3

A jogger runs one full 400 m lap in 100 s, finishing where she started. Find her average speed and her average velocity.
Answer and explanation

average speed 4 m/s; average velocity 0 m/s

Average speed = total distance ÷ time = 400 ÷ 100 = 4 m/s. Average velocity = displacement ÷ time = 0 ÷ 100 = 0 m/s, because she returns to the start.

Problem 4

A car's velocity increases from 8 m/s to 20 m/s in 4.0 s. Find its acceleration.
Answer and explanation

3 m/s²

a = (v − v₀) ÷ t = (20 − 8) ÷ 4 = 12 ÷ 4 = 3 m/s².

Problem 5

A stone is dropped from rest and falls freely for 2.0 s. Find its speed. (g = 9.8 m/s²)
Answer and explanation

19.6 m/s

v = v₀ + gt = 0 + 9.8 × 2.0 = 19.6 m/s.

Problem 6

Starting from rest, a train accelerates at 2.5 m/s² for 8.0 s. Find its final velocity.
Answer and explanation

20 m/s

v = v₀ + at = 0 + (2.5)(8.0) = 20 m/s.

Problem 7

For that same train (from rest, 2.5 m/s² for 8.0 s), how far does it travel?
Answer and explanation

80 m

Δx = v₀t + ½at² = 0 + ½(2.5)(8.0²) = ½(2.5)(64) = 80 m.

Problem 8

A car traveling at 25 m/s brakes uniformly to a stop in 5.0 s. Find (a) its acceleration and (b) the distance it travels while stopping.
Answer and explanation

(a) −5 m/s² (b) 62.5 m

(a) a = (0 − 25) ÷ 5 = −5 m/s². (b) Using the average-velocity form, Δx = ½(v₀ + v)t = ½(25 + 0)(5.0) = 62.5 m.

Problem 9

A motorcycle traveling at 30 m/s brakes at a constant −6.0 m/s². Find its stopping distance.
Answer and explanation

75 m

Use v² = v₀² + 2aΔx with v = 0: 0 = 30² + 2(−6.0)Δx, so 900 = 12Δx and Δx = 75 m.

Problem 10

A ball is thrown straight up at 19.6 m/s. Find (a) the time to reach its highest point and (b) its maximum height above the launch point. (g = 9.8 m/s²)
Answer and explanation

(a) 2.0 s (b) 19.6 m

(a) At the top the velocity is 0, so t = (v − v₀) ÷ a = (0 − 19.6) ÷ (−9.8) = 2.0 s. (b) Maximum height = v₀² ÷ (2 g) = 19.6² ÷ (2 × 9.8) = 384.16 ÷ 19.6 = 19.6 m.

Problem 11

A ball is thrown horizontally at 10 m/s from a cliff 44.1 m high. Find (a) its time in the air and (b) how far from the base it lands. (g = 9.8 m/s²)
Answer and explanation

(a) 3.0 s (b) 30 m

(a) Vertical motion: 44.1 = ½(9.8)t² = 4.9t², so t² = 9.0 and t = 3.0 s. (b) Horizontal motion is constant velocity: x = (10)(3.0) = 30 m.

Problem 12

A projectile is launched at 40 m/s at an angle of 30° above the horizontal. Find (a) the horizontal and (b) the vertical components of its initial velocity.
Answer and explanation

(a) ≈ 34.6 m/s (b) 20 m/s

(a) Horizontal component = v₀cosθ = 40 × cos 30° = 40 × 0.866 ≈ 34.6 m/s. (b) Vertical component = v₀sinθ = 40 × sin 30° = 40 × 0.5 = 20 m/s.

Problem 13

A ball is launched at 20 m/s at 45° above the horizontal on level ground. Find its range. (g = 9.8 m/s²)
Answer and explanation

≈ 40.8 m

Range = v₀²sin(2θ) ÷ g = (20²)(sin 90°) ÷ 9.8 = 400 × 1 ÷ 9.8 ≈ 40.8 m. A 45° launch gives the maximum range because sin(2θ) = 1 there.

Problem 14

Walk 5 m east then 3 m west. Give distance and displacement.
Answer and explanation

Distance 8 m; displacement+2 m with east positive.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 15

Chapter review: A student walks 12 m north in 6 s, then 4 m south in 4 s. Find average speed and average velocity, taking north positive.
Answer and explanation

Distance=16 m and elapsed time=10 s, so average speed=1.6 m/s. Displacement=+8 m, so average velocity=+0.8 m/s. Do not average the segment speeds without time weighting.

Distance=16 m and elapsed time=10 s, so average speed=1.6 m/s.

Displacement=+8 m, so average velocity=+0.8 m/s.

Do not average the segment speeds without time weighting.

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.

Problem 16

v changes−2 to−8 m/s in 3 s. Find acceleration and speed change.
Answer and explanation

a=(−8+2)/3=−2 m/s²; speed increases because velocity and acceleration point together.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 17

Chapter review: A cart changes from +8 to −4 m/s in 6 s under constant acceleration. Find acceleration, reversal time and displacement.
Answer and explanation

a=(−4−8)/6=−2 m/s². v=0 at 4 s. Displacement=8×6+½(−2)×6²=12 m; positive net displacement does not mean it moved east for the entire interval.

a=(−4−8)/6=−2 m/s². v=0 at 4 s.

Displacement=8×6+½(−2)×6²=12 m; positive net displacement does not mean it moved east for the entire interval.

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.

Problem 18

A velocity-time graph rises linearly from 0 to 6 m/s in 4 s. Find displacement.
Answer and explanation

Area of triangle=.5×4×6=12 m; acceleration slope=1.5 m/s².

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 19

Chapter review: 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.

Problem 20

A horizontal launch lasts 2 s at vx=3 m/s. Find horizontal range.
Answer and explanation

Δx=vxt=6 m. Vertical fall is½gt²≈19.6 m withg=9.8 m/s².

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 21

Chapter review: A ball is launched horizontally at 8.0 m/s from a 19.6 m ledge. Ignore drag and use g=9.8 m/s². Find flight time, horizontal range and landing velocity components.
Answer and explanation

0=19.6−4.9t² gives t=2.00 s. Range=8×2=16.0 m. Landing velocity=(8.0,−19.6) m/s in right/up-positive axes; speed≈21.2 m/s.

0=19.6−4.9t² gives t=2.00 s.

Range=8×2=16.0 m.

Landing velocity=(8.0,−19.6) m/s in right/up-positive axes; speed≈21.2 m/s.

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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