Motion & Kinematics — Chapter Practice, Grade 12

13 km

Distance is total path length: 8 + 5 = 13 km.

3 km

Displacement combines opposite directions: 8 − 5 = 3 km east.

displacement

Displacement has both magnitude and direction, making it a vector. Distance, speed, and mass are scalars.

Its displacement is zero but its distance equals the track's length

Returning to the start makes the displacement zero, while the distance traveled equals the full length of the lap.

13 m

The legs are perpendicular, so displacement = √(12² + 5²) = √169 = 13 m.

Velocity is always measured relative to a frame of reference. In the train's frame, the seated passenger isn't moving, so her velocity is zero. In the platform's frame, she is carried along with the train at, say, 100 km/h. Both are correct within their own frame because there is no absolute resting point in the universe — motion is relative to whatever origin you measure from.

Final minus initial position should match the signed path sum.

Distance 400 m; displacement 0 m.

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

Coordinates depend on the chosen reference frame; changes can then be interpreted consistently.

No. It supplies totals and duration, not the full motion history between events.

An average rate divides a defined change or amount by elapsed time. Average speed is total distance divided by duration; average velocity is displacement divided by duration. The same 4-second journey can therefore have 2 m/s average speed but only +0.5 m/s average velocity. Averaging two segment speeds by ordinary arithmetic works only for equal segment times. For unequal times, use total distance and total duration or the appropriate time-weighted calculation.

No. It supplies totals and duration, not the full motion history between events.

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.

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.

8 m/s

Average velocity = displacement ÷ time = 100 ÷ 12.5 = 8 m/s.

3 m/s²

a = (v − v₀) ÷ t = (22 − 10) ÷ 4 = 12 ÷ 4 = 3 m/s².

9.8 m/s² directed downward

Gravity acts the whole time, so the acceleration stays 9.8 m/s² downward even when the velocity is instantaneously zero.

49 m/s

v = v₀ + gt = 0 + 9.8 × 5.0 = 49 m/s.

-8 m/s²

a = (v − v₀) ÷ t = (4 − 28) ÷ 3 = −24 ÷ 3 = −8 m/s². Here the acceleration is opposite the stated positive velocity, so speed decreases during the interval. A negative acceleration does not mean slowing in every motion case.

Average velocity is the total displacement divided by the total time for the whole trip — it smooths everything into one number, like driving 270 km north in 3 hours for an average velocity of 90 km/h north. Instantaneous velocity is how fast (and which way) you are moving at one specific moment, like the 120 km/h your speedometer reads on the open highway or the 0 km/h at a rest stop. The average can hide big swings in the instantaneous value.

Acceleration sign alone does not say whether speed increases.

(15−3)/4=3 m/s².

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

Yes. A return trip has zero displacement but nonzero traveled distance.

Yes. The endpoint average is still +2 m/s², while instantaneous acceleration could vary.

Average acceleration is Δv/Δt, so the numerator is final velocity minus initial velocity with signs retained. Here the change from −3 to +5 m/s is +8 m/s over 4 s, giving +2 m/s². This average does not by itself prove acceleration was constant within the interval. The units mean a velocity change of meters per second for every second, not a distance traveled per second squared.

Yes. The endpoint average is still +2 m/s², while instantaneous acceleration could vary.

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.

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.

30 m

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

12 m/s

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

24 m

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

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.

40 m

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

(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.

These equations require constant acceleration over the interval.

Δx=½×2×25=25 m.

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

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

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.

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.

2 m

Horizontal motion is constant velocity: x = (4)(0.5) = 2 m.

stays constant

With no horizontal force acting, Newton's first law keeps the horizontal velocity constant for the whole flight.

3 s

Vertical motion: 44.1 = ½(9.8)t² = 4.9t², so t² = 9.0 and t = 3.0 s.

both land at the same time

Vertical motion is independent of horizontal motion, so both bullets are in identical free fall and land together.

80 m

First find the time: 78.4 = 4.9t², so t² = 16 and t = 4.0 s. Then range = (20)(4.0) = 80 m.

Once both are in the air, the dart and the monkey fall with the same downward acceleration g. In the time the dart travels to the monkey's original spot, gravity pulls the dart below its straight-line aim by exactly ½gt² — the same distance the monkey has fallen in that time. Because both drop by the identical amount, they meet. The dart's speed only changes when they meet, not whether they meet.

Solve vertical motion for time before using horizontal speed.

45=5t² gives t=3 s; range=18 m.

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

No. Vertical velocity is momentarily zero, but horizontal velocity remains nonzero without drag.

No. The vertical equation is unchanged; range doubles while the same flight time applies.

The two components have separate equations but share the same elapsed time. Once the vertical equation gives the later launch-level landing at about 3.27 s, horizontal displacement is 12×3.27≈39.2 m. Recombining the landing components gives speed √(12²+16²)=20 m/s. This equality with launch speed follows equal height and negligible drag; it is not a universal rule for every landing elevation or real ball.

No. The vertical equation is unchanged; range doubles while the same flight time applies.

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.

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.