Motion & Kinematics — Chapter quiz · Physics & Environmental Science, Grade 12
Describe motion with graphs, vectors, and bounded constant-acceleration models.
Problem 1
Answer and explanation
0 km
She ends where she began, so her displacement is 0 km (even though she traveled 12 km of distance).
Problem 2
- acceleration
- distance
- velocity
- displacement
Answer and explanation
distance
Distance has magnitude only, making it a scalar. Velocity, displacement, and acceleration are all vectors.
Problem 3
Answer and explanation
4 m/s²
a = (v − v₀) ÷ t = (25 − 5) ÷ 5 = 20 ÷ 5 = 4 m/s².
Problem 4
Answer and explanation
29.4 m/s
v = v₀ + gt = 0 + 9.8 × 3.0 = 29.4 m/s.
Problem 5
- 4.9 m/s² downward
- zero
- 9.8 m/s² downward
- 9.8 m/s² upward
Answer and explanation
9.8 m/s² downward
Gravity acts throughout the flight, so the acceleration is 9.8 m/s² downward even when the velocity is instantaneously zero.
Problem 6
Answer and explanation
20 m/s
v = v₀ + at = 0 + (4.0)(5.0) = 20 m/s.
Problem 7
Answer and explanation
50 m
Δx = v₀t + ½at² = 0 + ½(4.0)(5.0²) = ½(4.0)(25) = 50 m.
Problem 8
- displacement
- velocity
- distance
- acceleration
Answer and explanation
acceleration
Slope on a velocity-time graph is the change in velocity over time — that is, the acceleration.
Problem 9
Answer and explanation
96 m
Use v² = v₀² + 2aΔx with v = 0: 0 = 24² + 2(−3.0)Δx, so 576 = 6Δx and Δx = 96 m.
Problem 10
- stays constant
- decreases
- becomes zero at the top
- increases
Answer and explanation
stays constant
No horizontal force acts on the projectile, so its horizontal velocity stays constant for the whole flight.
Problem 11
Answer and explanation
36 m
Time to fall: 19.6 = 4.9t², so t² = 4.0 and t = 2.0 s. Range = (18)(2.0) = 36 m.
Problem 12
Answer and explanation
The two motions are independent: gravity only acts vertically, so the vertical motion is free fall (accelerating at g), while the horizontal motion has no force on it and stays at constant velocity. They share only the time in the air, so you can solve each as its own one-dimensional problem and link them through t.
Problem 13
Answer and explanation
Distance 8 m; displacement+2 m with east positive. Distance traveled and displacement can differ even on a straight line.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 14
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 15
Answer and explanation
a=(−8+2)/3=−2 m/s²; speed increases because velocity and acceleration point together. Negative acceleration need not mean slowing; compare its direction with velocity.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 16
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 17
Answer and explanation
Area of triangle=.5×4×6=12 m; acceleration slope=1.5 m/s². Constant-acceleration formulas do not apply across arbitrary changing acceleration.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 18
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 19
Answer and explanation
Δx=vxt=6 m. Vertical fall is½gt²≈19.6 m withg=9.8 m/s². Horizontal motion does not cancel gravity; independence assumes negligible air resistance.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 20
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.