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Kinematics · Solutions运动学 · 详解

Companion to the Practice Set · Mark-by-mark walkthroughs · AP-Feeder / ON / BC / AB styles练习题配套答案 · 逐分讲解 · AP 衔接 / 安 / 卑 / 阿省考风格

EASY易 MEDIUM中 HARD难 🇺🇸 US美 🇨🇦 ON安 🇨🇦 BC卑 🇨🇦 AB阿 AP-style MCQAP 风格选择题 AP-feeder FRQAP 衔接简答题 ON Provincial-style安大略省考风格 BC Provincial-style卑诗省考风格 AB Diploma-style阿尔伯塔毕业考风格 Honors荣誉级
PART I  ·  SHORT RESPONSE · SOLUTIONS第一部分  ·  短答题 · 详解AP MCQ + ON/BC short answer · 18 marksAP 选择题 + 安/卑省考短答 · 共 18 分

Section A · Worked SolutionsA 部分 · 详细解答

Q1EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §1 Displacement位移 · HS-PS2-1 [3 marks][3 分]

A hiker walks 40 m east then 30 m west. Magnitude of displacement?徒步者先东走 40 m，再西走 30 m，位移大小为多少？

Answer:答案：  (A)  $10\ \text{m}$

(a) Set up a sign convention and find net displacement设定正方向，求净位移 M1·A1·A1

Take east as positive. The two legs are $+40$ m and $-30$ m. Net displacement:取向东为正方向。两段路程分别为 $+40$ m 和 $-30$ m。净位移： $$ \Delta x \;=\; +40 + (-30) \;=\; +10 \;\text{m}. $$ Magnitude is $10$ m, option (A). The total distance walked is $40 + 30 = 70$ m (option B), a common trap.大小为 $10$ m，选 (A)。总路程为 $40 + 30 = 70$ m（选项 B），是常见陷阱。
Why the distractors fail.干扰项分析。
(B) $70\ \text{m}$: this is the total distance (scalar), not displacement (vector).这是总路程（标量），而非位移（矢量）。
(C) $50\ \text{m}$: comes from averaging the two legs or treating them as sides of a right triangle.对两段取平均或当作直角三角形两边处理所得。
(D) $35\ \text{m}$: average of 40 and 30.40 与 30 的平均值。
Displacement is a vector; distance is a scalar.位移是矢量；路程是标量。 Displacement asks "how far and in what direction from start to finish?" It is the straight-line arrow from the initial to the final position. Distance is the total path length, always non-negative. For motion along a straight line, displacement $= \Delta x = x_{\text{final}} - x_{\text{initial}}$. When the direction reverses, displacement and distance diverge. Locking in a sign convention (east $= +$) before calculating prevents sign errors and makes the vector nature explicit.位移问的是"从起点到终点有多远、朝哪个方向"，是从初位置到末位置的有向线段。路程是行走的总长度，恒非负。对于直线运动，位移 $= \Delta x = x_{\text{末}} - x_{\text{初}}$。当运动方向逆转时，位移与路程就会不同。动笔前先锁定正方向约定（东 $= +$），可防止符号错误，也让矢量性质更加清晰。
Q2EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §2 Average velocity平均速度 · HS-PS2-1 [3 marks][3 分]

Jogger: position changes from $+12$ m to $+92$ m in $20$ s. Average velocity?慢跑者：位置在 $20$ s 内从 $+12$ m 变为 $+92$ m。平均速度为多少？

Answer:答案：  (A)  $4\ \text{m/s}$

(a) Apply the average-velocity formula套用平均速度公式 M1·A1·A1

$$ \bar{v} \;=\; \frac{\Delta x}{\Delta t} \;=\; \frac{x_{\text{f}} - x_{\text{i}}}{\Delta t} \;=\; \frac{92 - 12}{20} \;=\; \frac{80}{20} \;=\; 4 \;\text{m/s.} $$ Option (A).选 (A)。
Why the distractors fail.干扰项分析。
(B) $4.6\ \text{m/s}$: divides the average of the two positions $(92 + 12)/2 = 52$ by 11 (arbitrary error).用两位置的平均值 $(92+12)/2=52$ 除以某数。
(C) $5.2\ \text{m/s}$: uses the final position $92$ divided by $\Delta t = 20 - 4 = 16$ (wrong denominator).错误地用末位置除以错误的时间间隔。
(D) $0.25\ \text{m/s}$: inverts $\Delta t / \Delta x$.把公式颠倒，算成 $\Delta t / \Delta x$。
Average velocity uses displacement, not distance.平均速度用位移，而非路程。 $\bar{v} = \Delta x / \Delta t$ is the ratio of the displacement to the time interval. The starting position $+12$ m is a clue that the origin is not at the jogger's starting point. The trap is to use only the final position $92$ m as if it were displacement. Always form $\Delta x = x_{\text{f}} - x_{\text{i}}$ explicitly before dividing. Average speed uses total distance $/ \Delta t$ and is always $\ge |\bar{v}|$; here they are equal because motion is one-directional.$\bar{v} = \Delta x / \Delta t$ 是位移与时间间隔之比。初位置 $+12$ m 提示原点不在慢跑者的出发点。常见陷阱是直接用末位置 $92$ m 代替位移。务必先明确写出 $\Delta x = x_{\text{末}} - x_{\text{初}}$，再相除。平均速率 $=$ 总路程 $/\Delta t$，始终 $\ge |\bar{v}|$；此题两者相等，因为运动方向未改变。
Q3EASY易 🇨🇦 ON安 ON Provincial-style安大略省考风格 §3 Acceleration加速度 · SPH3U B2 [4 marks][4 分]

Car: $12$ m/s to $30$ m/s in $6.0$ s. (a) Acceleration. (b) Sign interpretation. (c) Assumption.汽车：$12$ m/s 加速至 $30$ m/s，历时 $6.0$ s。(a) 加速度。(b) 符号含义。(c) 假设条件。

Answer:答案：  (a) $a = 3.0\ \text{m/s}^2$  ·  (b) positive: speeding up in the direction of motion正值：沿运动方向加速  ·  (c) constant (uniform) acceleration加速度恒定（匀加速）

(a) Compute acceleration with units计算加速度（带单位） M1·A1

$$ a \;=\; \frac{\Delta v}{\Delta t} \;=\; \frac{30 - 12}{6.0} \;=\; \frac{18}{6.0} \;=\; 3.0 \;\text{m/s}^2. $$

(b) Interpret the sign解释符号含义 A1

The positive value means the velocity is increasing. Since the car is moving in the positive direction and speeding up, the acceleration is in the same direction as the motion.正值说明速度在增大。汽车沿正方向行驶且在加速，因此加速度与运动方向相同。

(c) State the assumption写出假设条件 A1

The calculation $a = \Delta v / \Delta t$ gives the average acceleration over the interval. To use this as the (constant) acceleration, we must assume the acceleration is uniform (constant) throughout the 6.0 s.公式 $a = \Delta v / \Delta t$ 给出的是该时间间隔内的平均加速度。要将其视为（恒定）加速度，必须假设 $6.0$ s 内加速度均匀（恒定）。
$\Delta v / \Delta t$ gives average acceleration; uniform motion makes average $=$ instantaneous.$\Delta v / \Delta t$ 给出平均加速度；匀加速时平均值等于瞬时值。 In everyday speech "the car accelerated from 12 to 30 m/s in 6 s" implies uniform acceleration, but in physics this is an assumption that must be stated. A real car has varying engine output and drag; only under the uniformity assumption can we use $a = \Delta v / \Delta t$ as the instantaneous acceleration and then apply SUVAT equations. Provincial examiners award the (c) mark specifically to test whether students distinguish average from instantaneous quantities.日常语言中"汽车在 6 s 内从 12 m/s 加速到 30 m/s"暗示匀加速，但物理上这是需明确写出的假设。真实汽车的发动机输出和阻力是变化的；只有在匀加速假设下，才能把 $a = \Delta v / \Delta t$ 视为瞬时加速度并进而使用 SUVAT 方程。省考阅卷人设置 (c) 小问，正是专门考查学生能否区分平均量与瞬时量。
Q4MEDIUM中 🇺🇸 US美 AP-style MCQAP 风格选择题 §5 Motion graphs运动图像 · HS-PS2-1 [3 marks][3 分]

$v$-$t$ graph: straight line from $(0\ \text{s}, 4\ \text{m/s})$ to $(8\ \text{s}, 20\ \text{m/s})$. Acceleration?$v$-$t$ 图：从 $(0\ \text{s}, 4\ \text{m/s})$ 到 $(8\ \text{s}, 20\ \text{m/s})$ 的直线。加速度为多少？

Answer:答案：  (A)  $2\ \text{m/s}^2$

(a) Slope of the $v$-$t$ graph $=$ acceleration$v$-$t$ 图像的斜率 $=$ 加速度 M1·A1·A1

$$ a \;=\; \text{slope} \;=\; \frac{\Delta v}{\Delta t} \;=\; \frac{20 - 4}{8 - 0} \;=\; \frac{16}{8} \;=\; 2 \;\text{m/s}^2. $$ Option (A).选 (A)。
Why the distractors fail.干扰项分析。
(B) $2.5\ \text{m/s}^2$: uses $20 / 8$ without subtracting the initial velocity.用 $20/8$ 而未减去初速度。
(C) $16\ \text{m/s}^2$: gives $\Delta v$ alone without dividing by $\Delta t$.只算出 $\Delta v$，未除以 $\Delta t$。
(D) $0.5\ \text{m/s}^2$: inverts the ratio, $\Delta t / \Delta v$.将比值颠倒为 $\Delta t / \Delta v$。
On a $v$-$t$ graph: slope $=$ acceleration, area $=$ displacement.在 $v$-$t$ 图中：斜率 $=$ 加速度，面积 $=$ 位移。 Two key readings from a $v$-$t$ graph: (1) the gradient of the line gives the acceleration (rise over run in velocity-over-time units); (2) the area under the line gives the displacement. A common error is to confuse the value of $v$ at a point with the change in $v$. Here the object already has $4$ m/s at $t = 0$; the acceleration is set by how much $v$ changes, not by what $v$ is. If the initial velocity were zero, options (A) and (B) would give the same answer. The non-zero initial velocity is the discriminator.从 $v$-$t$ 图可做两种关键读取：(1) 直线的斜率给出加速度（速度随时间的变化率）；(2) 直线下方的面积给出位移。常见错误是把某点处 $v$ 的值与 $v$ 的变化量混淆。此题 $t = 0$ 时物体已有 $4$ m/s；加速度由 $v$ 变化了多少决定，而非 $v$ 的大小。若初速度为零，选项 (A) 与 (B) 会给出相同答案：正是非零初速度充当了区分器。
Q5MEDIUM中 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §4 SUVAT匀变速方程 · Physics 11 [5 marks][5 分]

Cyclist decelerates from $12$ m/s to rest over $48$ m. (a) Acceleration. (b) Time. (c) Assumption.骑行者从 $12$ m/s 匀减速至静止，经过 $48$ m。(a) 加速度。(b) 时间。(c) 假设条件。

Answer:答案：  (a) $a = -1.5\ \text{m/s}^2$  ·  (b) $t = 8.0\ \text{s}$  ·  (c) constant (uniform) acceleration恒定（匀变速）加速度

(a) Find acceleration using $v^2 = u^2 + 2as$用 $v^2 = u^2 + 2as$ 求加速度 M1·A1

Known: $u = 12$ m/s, $v = 0$, $s = 48$ m. Substituting:已知：$u = 12$ m/s，$v = 0$，$s = 48$ m。代入： $$ 0 \;=\; 12^2 + 2a(48) \;\Longrightarrow\; 2a(48) \;=\; -144 \;\Longrightarrow\; a \;=\; -1.5 \;\text{m/s}^2. $$ The negative sign confirms deceleration (acceleration opposite to motion).负号确认为减速（加速度方向与运动方向相反）。

(b) Find time using $v = u + at$用 $v = u + at$ 求时间 M1·A1

$$ 0 \;=\; 12 + (-1.5)t \;\Longrightarrow\; t \;=\; \frac{12}{1.5} \;=\; 8.0 \;\text{s}. $$

(c) State the assumption写出假设条件 A1

The SUVAT equations are valid only if acceleration is constant (uniform) throughout the motion. Real braking involves changing friction forces, but the problem states "uniformly," which licenses the assumption.SUVAT 方程仅在加速度全程恒定（匀变速）时成立。实际刹车时摩擦力会变化，但题目已说明"匀减速"，因此该假设成立。
Choose your SUVAT equation by listing knowns and the single unknown.列出已知量和唯一未知量，再选对应的 SUVAT 方程。 The five SUVAT variables are $s, u, v, a, t$. Part (a) gives $u, v, s$ and asks for $a$, pointing to $v^2 = u^2 + 2as$ (no $t$ needed). Part (b) then gives $u, v, a$ and asks for $t$, pointing to $v = u + at$. This two-step strategy is always safe: solve whichever equation contains three knowns and the one unknown, avoid introducing extra variables. The negative value of $a$ is expected and correct for deceleration. Never drop the sign: it carries directional information.SUVAT 五个变量为 $s, u, v, a, t$。(a) 已知 $u, v, s$，求 $a$，对应 $v^2 = u^2 + 2as$（无需 $t$）。(b) 已知 $u, v, a$，求 $t$，对应 $v = u + at$。这一两步策略始终安全：找含三个已知量和一个未知量的方程，避免引入多余变量。$a$ 为负是减速的正确结果：绝对不要丢掉负号，它携带方向信息。
PART II  ·  EXTENDED RESPONSE · SOLUTIONS第二部分  ·  简答题 · 详解AP-feeder FRQ + honors · 35 marksAP 衔接简答题 + 荣誉级 · 共 35 分

Section B · Worked SolutionsB 部分 · 详细解答

Q6MEDIUM中 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §4 SUVAT (two-phase)匀变速方程（两阶段） · HS-PS2-1 [8 marks][8 分]

Car: starts from rest, $a = 2.5\ \text{m/s}^2$ for $8.0$ s, then constant velocity for $12$ s. (a) Velocity after 8 s. (b) Distance phase 1. (c) Distance phase 2. (d) Average velocity.汽车：从静止以 $a = 2.5\ \text{m/s}^2$ 加速 $8.0$ s，再以匀速行驶 $12$ s。(a) 8 s 后速度。(b) 第一阶段距离。(c) 第二阶段距离。(d) 平均速度。

Answer:答案：  (a) $v = 20\ \text{m/s}$  ·  (b) $s_1 = 80\ \text{m}$  ·  (c) $s_2 = 240\ \text{m}$  ·  (d) $\bar{v} = 16\ \text{m/s}$

(a) Velocity after 8.0 s8.0 s 后的速度 M1·A1

$$ v \;=\; u + at \;=\; 0 + 2.5 \times 8.0 \;=\; 20 \;\text{m/s.} $$

(b) Distance during phase 1 (acceleration)第一阶段（加速段）距离 M1·A1

$$ s_1 \;=\; ut + \tfrac{1}{2}at^2 \;=\; 0 + \tfrac{1}{2}(2.5)(8.0)^2 \;=\; \tfrac{1}{2}(2.5)(64) \;=\; 80 \;\text{m.} $$

(c) Distance during phase 2 (constant velocity)第二阶段（匀速段）距离 M1·A1

$$ s_2 \;=\; v \times t \;=\; 20 \times 12 \;=\; 240 \;\text{m.} $$

(d) Average velocity over 20 s20 s 内的平均速度 M1·A1

$$ \bar{v} \;=\; \frac{s_1 + s_2}{\Delta t} \;=\; \frac{80 + 240}{20} \;=\; \frac{320}{20} \;=\; 16 \;\text{m/s.} $$
Average velocity over two phases requires total displacement over total time, not the average of the two velocities.两阶段平均速度须用总位移除以总时间，而非两段速度的平均值。 A common trap: $(0 + 20)/2 = 10$ m/s for phase 1, $(20 + 20)/2 = 20$ m/s for phase 2, then averaging to $15$ m/s. This is wrong because the two phases last different distances. The only correct method is $\bar{v} = \Delta x_{\text{total}} / \Delta t_{\text{total}}$. Check: $16 \times 20 = 320$ m $= 80 + 240$. Notice also that within phase 1 alone the average velocity is $(0 + 20)/2 = 10$ m/s (valid for uniform acceleration), giving $s_1 = 10 \times 8 = 80$ m, consistent with the SUVAT result.常见陷阱：第一阶段 $(0 + 20)/2 = 10$ m/s，第二阶段 $(20 + 20)/2 = 20$ m/s，再平均得 $15$ m/s。这是错的，因为两段持续时间不同。唯一正确的方法是 $\bar{v} = \Delta x_{\text{总}} / \Delta t_{\text{总}}$。验证：$16 \times 20 = 320$ m $= 80 + 240$，$\checkmark$。还要注意：仅在第一阶段内，平均速度为 $(0+20)/2 = 10$ m/s（对匀加速有效），给出 $s_1 = 10 \times 8 = 80$ m，与 SUVAT 结果一致。
Q7MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §6 Free fall自由落体 · SPH3U B3 [8 marks][8 分]

Ball thrown straight up at $24$ m/s from ground. $g = 9.8\ \text{m/s}^2$. (a) Max height. (b) Time to top. (c) Velocity on return. (d) Total flight time.球以 $24$ m/s 从地面竖直上抛，$g = 9.8\ \text{m/s}^2$。(a) 最大高度。(b) 到达最高点的时间。(c) 返回时速度。(d) 总飞行时间。

Answer:答案：  (a) $h = 29.4\ \text{m}$  ·  (b) $t = 2.45\ \text{s}$  ·  (c) $v = -24\ \text{m/s}$ (downward)（向下）  ·  (d) $T = 4.9\ \text{s}$

(a) Maximum height using $v^2 = u^2 - 2gh$用 $v^2 = u^2 - 2gh$ 求最大高度 M1·A1·A1

Take upward as positive. At the top, $v = 0$. With $u = +24$ m/s, $a = -9.8\ \text{m/s}^2$:取向上为正方向。在最高点 $v = 0$。已知 $u = +24$ m/s，$a = -9.8\ \text{m/s}^2$： $$ 0 \;=\; (24)^2 - 2(9.8)h \;\Longrightarrow\; h \;=\; \frac{576}{19.6} \;=\; 29.4 \;\text{m.} $$

(b) Time to reach the top到达最高点的时间 M1·A1

$$ v \;=\; u + at \;\Longrightarrow\; 0 \;=\; 24 - 9.8\,t \;\Longrightarrow\; t \;=\; \frac{24}{9.8} \;=\; 2.45 \;\text{s.} $$

(c) Velocity on return to launch height返回抛出高度时的速度 M1·A1

By symmetry of free fall, the speed on return equals the launch speed, but the direction is reversed:由自由落体的对称性，返回时速率与抛出时相同，但方向相反： $$ v \;=\; -24 \;\text{m/s} \quad (\text{downward}). $$

(d) Total flight time总飞行时间 A1

$$ T \;=\; 2 \times 2.45 \;=\; 4.9 \;\text{s.} $$
Symmetry of free-fall: up-time equals down-time, launch speed equals landing speed.自由落体对称性：上升时间等于下降时间，抛出速率等于落回速率。 For any object launched vertically and returning to the same height (no air resistance): (1) time up $=$ time down, so total flight time $= 2t_{\text{up}}$; (2) speed at the original height on the way down $=$ launch speed. This symmetry is a direct consequence of the parabolic $y$-$t$ profile being symmetric about its vertex. State your sign convention before calculating; the "return velocity is $-24$ m/s" answer is incomplete without noting that up is positive. A common error is to report only the magnitude $24$ m/s and forget the direction, losing the A1 for this part on ON exams.对于任何竖直抛出后返回同一高度的物体（无空气阻力）：(1) 上升时间 $=$ 下降时间，总飞行时间 $= 2t_{\text{上}}$；(2) 下落至原高度时的速率 $=$ 初速率。该对称性是 $y$-$t$ 图像抛物线关于顶点对称的直接结果。计算前先声明正方向约定；"返回速度为 $-24$ m/s"这一答案若不注明向上为正，则不完整。常见错误是只报速率大小 $24$ m/s 而忘记方向，在安大略省考中会丢掉该小问的 A1。
Q8HARD难 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §5 Motion graphs运动图像 · Physics 11 [9 marks][9 分]

Train $v$-$t$ graph: 3 segments (0-10 s: 0 to 25 m/s; 10-30 s: constant; 30-40 s: 25 to 0 m/s). (a) Accelerations. (b) Total displacement. (c) Average velocity. (d) Description.火车 $v$-$t$ 图三段（0-10 s：0 至 25 m/s；10-30 s：匀速；30-40 s：25 至 0 m/s）。(a) 各段加速度。(b) 总位移。(c) 平均速度。(d) 描述运动。

Answer:答案：  (a) $+2.5,\ 0,\ -2.5\ \text{m/s}^2$  ·  (b) $750\ \text{m}$  ·  (c) $18.75\ \text{m/s}$  ·  (d) uniform acceleration, constant velocity, uniform deceleration to rest匀加速、匀速、匀减速至静止

(a) Acceleration in each phase (slope of $v$-$t$)各阶段加速度（$v$-$t$ 斜率） A1·A1·A1

Phase 1 ($0$ to $10$ s): $a_1 = (25-0)/(10-0) = +2.5\ \text{m/s}^2$.第一阶段（$0$ 到 $10$ s）：$a_1 = (25-0)/(10-0) = +2.5\ \text{m/s}^2$。
Phase 2 ($10$ to $30$ s): $a_2 = 0\ \text{m/s}^2$ (constant $v$).第二阶段（$10$ 到 $30$ s）：$a_2 = 0\ \text{m/s}^2$（匀速）。
Phase 3 ($30$ to $40$ s): $a_3 = (0-25)/(40-30) = -2.5\ \text{m/s}^2$.第三阶段（$30$ 到 $40$ s）：$a_3 = (0-25)/(40-30) = -2.5\ \text{m/s}^2$。

(b) Total displacement (area under $v$-$t$ graph)总位移（$v$-$t$ 图下方面积） M1·A1·A1

$$ s \;=\; \underbrace{\tfrac{1}{2}(10)(25)}_{\text{triangle 1}} + \underbrace{(20)(25)}_{\text{rectangle}} + \underbrace{\tfrac{1}{2}(10)(25)}_{\text{triangle 2}} \;=\; 125 + 500 + 125 \;=\; 750 \;\text{m.} $$

(c) Average velocity平均速度 M1·A1

$$ \bar{v} \;=\; \frac{750}{40} \;=\; 18.75 \;\text{m/s.} $$

(d) Description运动描述 A1

The train speeds up uniformly from rest to $25$ m/s, then travels at constant velocity, then slows uniformly to rest.火车从静止匀加速至 $25$ m/s，然后匀速行驶，最后匀减速至静止。
On a $v$-$t$ graph: slope $=$ acceleration (signed), area $=$ displacement (signed).在 $v$-$t$ 图中：斜率 $=$ 加速度（有方向），面积 $=$ 位移（有方向）。 For part (b), the area method is far faster than three separate SUVAT calculations. The shapes are a triangle, a rectangle, and another triangle; their areas add directly to the displacement. A negative area (if $v$ goes below zero) would indicate displacement in the negative direction. For part (c), the standard trap is to average the three phase velocities: $(12.5 + 25 + 12.5)/3 = 16.67$ m/s, which is wrong because the phases have unequal durations. Always use total displacement over total time for average velocity.对于 (b)，面积法比分段做三次 SUVAT 计算快得多。图形由三角形、矩形和另一个三角形构成，面积直接相加即为位移。若 $v$ 降至零以下，负面积表示负方向的位移。对于 (c)，标准陷阱是对三段速度取平均：$(12.5 + 25 + 12.5)/3 = 16.67$ m/s，这是错的，因为各段持续时间不同。平均速度必须始终用总位移除以总时间。
Q9HARD难Honors荣誉级 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §7 Projectile抛体运动 · HS-PS2-1 (above 1D floor)（超出一维基准） [10 marks][10 分]

Ball launched horizontally at $20$ m/s from cliff $44.1$ m high. $g = 9.8\ \text{m/s}^2$. (a) Time of flight. (b) Range. (c) Vertical velocity at impact. (d) Speed at impact. (e) Angle below horizontal.球从 $44.1$ m 高悬崖以 $20$ m/s 水平抛出，$g = 9.8\ \text{m/s}^2$。(a) 飞行时间。(b) 水平射程。(c) 落地竖直速度。(d) 落地速率。(e) 速度方向与水平方向夹角。

Answer:答案：  (a) $t = 3.0\ \text{s}$  ·  (b) $R = 60\ \text{m}$  ·  (c) $v_y = 29.4\ \text{m/s}\ \downarrow$  ·  (d) $v = 35.6\ \text{m/s}$  ·  (e) $\theta = 55.8°$

(a) Time of flight (vertical free fall)飞行时间（竖直自由落体） M1·A1

Horizontal launch means $v_{y,0} = 0$. Using $h = \tfrac{1}{2}g t^2$:水平抛出意味着 $v_{y,0} = 0$。使用 $h = \tfrac{1}{2}g t^2$： $$ 44.1 \;=\; \tfrac{1}{2}(9.8)t^2 \;\Longrightarrow\; t^2 \;=\; \frac{44.1}{4.9} \;=\; 9.0 \;\Longrightarrow\; t \;=\; 3.0 \;\text{s.} $$

(b) Horizontal range水平射程 M1·A1

$$ R \;=\; v_x \cdot t \;=\; 20 \times 3.0 \;=\; 60 \;\text{m.} $$

(c) Vertical velocity at impact落地竖直速度 M1·A1

$$ v_y \;=\; g\,t \;=\; 9.8 \times 3.0 \;=\; 29.4 \;\text{m/s} \;\; (\text{downward}). $$

(d) Speed at impact (magnitude of velocity vector)落地速率（速度矢量的大小） M1·A1

$$ v \;=\; \sqrt{v_x^2 + v_y^2} \;=\; \sqrt{20^2 + 29.4^2} \;=\; \sqrt{400 + 864.36} \;=\; \sqrt{1264.36} \;\approx\; 35.6 \;\text{m/s.} $$

(e) Angle below horizontal速度方向低于水平方向的角度 M1·A1

$$ \theta \;=\; \arctan\!\left(\frac{v_y}{v_x}\right) \;=\; \arctan\!\left(\frac{29.4}{20}\right) \;=\; \arctan(1.47) \;\approx\; 55.8°. $$
Projectile motion: horizontal and vertical components are independent, coupled only by time $t$.抛体运动：水平与竖直分量相互独立，仅通过时间 $t$ 耦合。 The key principle: gravity acts only on the vertical component; the horizontal component is unaffected (assuming no air resistance). This gives two separate 1D kinematics problems that share the same $t$. Step 1: find $t$ from the vertical equation (free fall). Step 2: use that $t$ in the horizontal equation (uniform motion). Step 3: combine $v_x$ (constant throughout) and $v_y$ (at impact) using the Pythagorean theorem for speed, and $\arctan(v_y/v_x)$ for the angle. A common sign-convention error is to set $g = -9.8$ and then lose a negative somewhere; it is safer to assign the downward direction as positive for the vertical sub-problem when the object only moves down.核心原理：重力仅作用于竖直分量；水平分量不受影响（假设无空气阻力）。这产生两个独立的一维运动问题，共享同一个 $t$。第一步：从竖直方程（自由落体）求 $t$。第二步：把该 $t$ 代入水平方程（匀速运动）。第三步：用勾股定理合成 $v_x$（全程恒定）与 $v_y$（落地时）得速率，用 $\arctan(v_y/v_x)$ 得角度。常见的正负号错误是令 $g = -9.8$ 后在某处丢掉负号；当物体只向下运动时，将向下定为竖直子问题的正方向更为安全。
PART III  ·  MODELING / APPLIED · SOLUTIONS第三部分  ·  建模与应用 · 详解AB Diploma + Universal · 28 marks阿省毕业考 + 通用题型 · 共 28 分

Section C · Modeling and ApplicationsC 部分 · 建模与应用

Q10MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §4 SUVAT (applied)匀变速方程（应用） · 20–A1.3k [9 marks][9 分]

Car at $30$ m/s brakes uniformly at $6.0\ \text{m/s}^2$. Reaction time $0.70$ s. (a) Braking distance. (b) Braking time. (c) Reaction-phase distance. (d) Total stopping distance.汽车以 $30$ m/s 行驶，以 $6.0\ \text{m/s}^2$ 匀减速刹车。反应时间 $0.70$ s。(a) 刹车距离。(b) 刹车时间。(c) 反应阶段距离。(d) 总停车距离。

Answer:答案：  (a) $75\ \text{m}$  ·  (b) $5.0\ \text{s}$  ·  (c) $21\ \text{m}$  ·  (d) $96\ \text{m}$

(a) Braking distance using $v^2 = u^2 + 2as$用 $v^2 = u^2 + 2as$ 求刹车距离 M1·A1·A1

$u = 30$ m/s, $v = 0$, $a = -6.0\ \text{m/s}^2$:$u = 30$ m/s，$v = 0$，$a = -6.0\ \text{m/s}^2$： $$ 0 \;=\; 30^2 + 2(-6.0)s \;\Longrightarrow\; 12s \;=\; 900 \;\Longrightarrow\; s \;=\; 75 \;\text{m.} $$

(b) Braking time using $v = u + at$用 $v = u + at$ 求刹车时间 M1·A1

$$ 0 \;=\; 30 + (-6.0)t \;\Longrightarrow\; t \;=\; \frac{30}{6.0} \;=\; 5.0 \;\text{s.} $$

(c) Reaction-phase distance (constant velocity)反应阶段距离（匀速） M1·A1

$$ s_{\text{reaction}} \;=\; u \cdot t_{\text{reaction}} \;=\; 30 \times 0.70 \;=\; 21 \;\text{m.} $$

(d) Total stopping distance总停车距离 M1·A1

$$ s_{\text{total}} \;=\; s_{\text{reaction}} + s_{\text{braking}} \;=\; 21 + 75 \;=\; 96 \;\text{m.} $$
Total stopping distance has two phases: the reaction phase (constant velocity) and the braking phase (deceleration).总停车距离分两阶段：反应阶段（匀速）和制动阶段（减速）。 This is a real-world safety application standard on AB Physics 20 diploma exams. The reaction distance $= u \cdot t_{\text{reaction}}$ is a simple $d = vt$ calculation, while the braking distance uses SUVAT with $a = -6.0\ \text{m/s}^2$. A common error is to forget the reaction distance entirely, reporting only $75$ m. At highway speeds, the reaction distance is significant: here it adds $21$ m, a $28\%$ increase. Also notice that doubling the initial speed quadruples the braking distance (since $s \propto u^2$), a key safety insight used in speed limit design.这是阿省物理 20 毕业考的标准实际应用题。反应距离 $= u \cdot t_{\text{反应}}$ 是简单的 $d = vt$ 计算，制动距离用 $a = -6.0\ \text{m/s}^2$ 的 SUVAT 方程。常见错误是完全忘记反应距离，只报 $75$ m。在高速公路速度下，反应距离不可忽视：此处额外增加 $21$ m，增幅达 $28\%$。还要注意：初速加倍后制动距离变为四倍（因为 $s \propto u^2$），这是制定限速标准时的关键安全依据。
Q11MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §6 Free fall (applied)自由落体（应用） · 20–A1.5k [9 marks][9 分]

Stone dropped from bridge, hits water in $2.5$ s, $g = 9.8\ \text{m/s}^2$. (a) Bridge height. (b) Impact speed. (c) Speed at $1.5$ s. (d) Distance in first $1.5$ s.石头从桥上落下，$2.5$ s 后落水，$g = 9.8\ \text{m/s}^2$。(a) 桥高。(b) 落水速率。(c) $1.5$ s 时速率。(d) 前 $1.5$ s 下落距离。

Answer:答案：  (a) $30.6\ \text{m}$  ·  (b) $24.5\ \text{m/s}$  ·  (c) $14.7\ \text{m/s}$  ·  (d) $11.0\ \text{m}$

(a) Height of bridge using $s = \tfrac{1}{2}g t^2$用 $s = \tfrac{1}{2}g t^2$ 求桥高 M1·A1·A1

$$ h \;=\; \tfrac{1}{2}(9.8)(2.5)^2 \;=\; \tfrac{1}{2}(9.8)(6.25) \;=\; 30.625 \;\approx\; 30.6 \;\text{m.} $$

(b) Speed at impact using $v = gt$用 $v = gt$ 求落水速率 M1·A1

$$ v \;=\; 9.8 \times 2.5 \;=\; 24.5 \;\text{m/s.} $$

(c) Speed at $t = 1.5$ s$t = 1.5$ s 时的速率 M1·A1

$$ v_{1.5} \;=\; 9.8 \times 1.5 \;=\; 14.7 \;\text{m/s.} $$

(d) Distance in first $1.5$ s前 $1.5$ s 内下落距离 M1·A1

$$ s_{1.5} \;=\; \tfrac{1}{2}(9.8)(1.5)^2 \;=\; \tfrac{1}{2}(9.8)(2.25) \;=\; 11.025 \;\approx\; 11.0 \;\text{m.} $$
For an object dropped from rest: $v = gt$ and $s = \tfrac{1}{2}gt^2$ are the two master equations.对于由静止落下的物体：$v = gt$ 和 $s = \tfrac{1}{2}gt^2$ 是两个核心方程。 When $u = 0$ (dropped from rest), the SUVAT equations simplify enormously. Speed grows linearly with time: $v = gt$. Distance grows quadratically: $s = \tfrac{1}{2}gt^2$. Check: at $t = 1.5$ s, $s = 11.0$ m (less than half the bridge height), confirming that the stone spends less than half its fall distance in the first 1.5 s out of 2.5 s. This is the "late rush" of free fall: objects cover more distance per second as they accelerate. The ratio of distances fallen in the first and second equal time intervals is always $1 : 3$ (Galileo's odd-number rule: 1, 3, 5, ... for equal time increments), which provides a rapid check for free-fall problems.当 $u = 0$（由静止落下）时，SUVAT 方程大幅简化。速度随时间线性增长：$v = gt$。距离按二次方增长：$s = \tfrac{1}{2}gt^2$。验证：$t = 1.5$ s 时 $s = 11.0$ m（不到桥高的一半），说明石头在 $2.5$ s 总时间里的前 $1.5$ s 只下落了不到一半的距离。这就是自由落体的"末段加速"：物体加速后每秒覆盖更多距离。前后两段等时间内落距之比永远是 $1 : 3$（伽利略奇数定律：等时间增量对应 $1, 3, 5, \ldots$），可用于快速核对自由落体问题。
Q12HARD难 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §4 + relative motion相对运动 · HS-PS2-1 [10 marks][10 分]

Car A at $25$ m/s constant. Car B starts from rest at same point, $a = 4.0\ \text{m/s}^2$. (a) Time B catches A. (b) Distance at catch. (c) B's velocity. (d) Explain why $v_B = 2v_A$.A 车匀速 $25$ m/s；B 车从同一点静止出发，$a = 4.0\ \text{m/s}^2$。(a) B 追上 A 的时间。(b) 追上时的距离。(c) B 的速度。(d) 解释为何 $v_B = 2v_A$。

Answer:答案：  (a) $t = 12.5\ \text{s}$  ·  (b) $312.5\ \text{m}$  ·  (c) $50\ \text{m/s}$  ·  (d) B's average velocity equals A's constant velocity (equal displacement, equal time)B 的平均速度等于 A 的匀速（位移相等，时间相等）

(a) Time for B to catch A (equating positions)B 追上 A 的时间（令位置相等） M1·A1·A1

Position of A: $x_A = 25t$. Position of B: $x_B = \tfrac{1}{2}(4.0)t^2 = 2t^2$. Set $x_A = x_B$:A 的位置：$x_A = 25t$。B 的位置：$x_B = \tfrac{1}{2}(4.0)t^2 = 2t^2$。令 $x_A = x_B$： $$ 25t \;=\; 2t^2 \;\Longrightarrow\; 2t^2 - 25t \;=\; 0 \;\Longrightarrow\; t(2t - 25) \;=\; 0. $$ Solutions: $t = 0$ (start) or $t = 12.5$ s (catch-up).解：$t = 0$（出发时刻）或 $t = 12.5$ s（追上时刻）。

(b) Distance at catch-up追上时的距离 M1·A1

$$ x_A(12.5) \;=\; 25 \times 12.5 \;=\; 312.5 \;\text{m.} $$

(c) B's velocity at catch-up追上时 B 的速度 M1·A1

$$ v_B \;=\; at \;=\; 4.0 \times 12.5 \;=\; 50 \;\text{m/s.} $$

(d) Why $v_B = 2 v_A$为何 $v_B = 2 v_A$ M1·A1·A1

B starts from rest and accelerates uniformly. Its average velocity over the interval $[0, t^{*}]$ isB 从静止匀加速出发，在 $[0, t^{*}]$ 区间内的平均速度为 $$ \bar{v}_B \;=\; \frac{0 + v_B}{2} \;=\; \frac{v_B}{2}. $$ For B to cover the same displacement as A in the same time $t^{*}$, we need $\bar{v}_B = v_A$. HenceB 要在相同时间 $t^{*}$ 内覆盖与 A 相同的位移，需要 $\bar{v}_B = v_A$。因此 $$ \frac{v_B}{2} \;=\; v_A \;\Longrightarrow\; v_B \;=\; 2\,v_A. $$ This is a consequence of the mean-value theorem for linear (uniform) acceleration: the average velocity equals the midpoint of the initial and final velocities.这是线性（匀）加速度情形下中值定理的推论：平均速度等于初末速度的中间值。
Average velocity reasoning: the catch-up condition forces $\bar{v}_B = v_A$, which directly yields $v_B = 2v_A$.平均速度推理：追及条件要求 $\bar{v}_B = v_A$，从而直接得出 $v_B = 2v_A$。 This result is universal for the "constant-velocity car A, accelerating-from-rest car B" catch-up scenario: B always reaches exactly twice A's speed at the moment of catch-up, regardless of the values of $v_A$ and $a_B$. The underlying reason is elegant: equal displacement in equal time forces equal average velocity; uniform acceleration from rest makes average velocity $= $ half the final velocity; therefore final velocity $= 2 \times$ constant velocity. This argument generalises: if B starts from $u_B > 0$ (not rest), the factor of 2 no longer holds, so the argument is specific to $u_B = 0$.这一结果对"A 车匀速、B 车从静止加速"追及场景具有普遍性：无论 $v_A$ 和 $a_B$ 取何值，B 追上 A 的瞬间速度永远恰好是 A 速度的两倍。背后的道理很优美：相同时间内相同位移要求平均速度相等；从静止匀加速时平均速度 $=$ 末速度的一半；因此末速度 $= 2 \times$ 匀速。注意该论证有特定条件：若 B 的初速度 $u_B > 0$（非从静止），则两倍关系不再成立，故该论证专门适用于 $u_B = 0$ 的情形。