Each item below restates the prompt and choices, marks the correct letter, and gives a one- to three-sentence justification keyed to the AP topic.以下每题均重述题干与选项，标出正确答案，并给出一至三句与 AP 考点对应的解析。
Q1EASY1.1 Scalars and Vectors1.1 标量与矢量No Calculator
Which pair lists one scalar and one vector, in that order?下列哪组依次列出了一个标量和一个矢量？
(A)velocity, speed速度，速率
(B)distance, displacement路程，位移
(C)displacement, distance位移，路程
(D)acceleration, force加速度，力
Answer:答案：(B)
Distance is a path length (scalar, sign-free); displacement is a vector with magnitude and direction. (A) reverses the order. (C) flips them. (D) lists two vectors.路程是路径长度（标量，无符号）；位移是有大小和方向的矢量。(A) 颠倒了顺序，(C) 将两者互换，(D) 列出的是两个矢量。
Insight.要点。Distinguish a scalar from a vector by one test: does the quantity need a direction to be complete? Distance only asks how far along the path, while displacement also asks in which direction. Apply that test before looking at the answer choices.区分标量与矢量只需一个检验：该量是否需要方向才算完整？路程只问沿路径走了多远，而位移还要问朝哪个方向。先做这个检验，再看选项。
A particle's position is $x(t) = 3t^2 - 12t + 5$ (SI units). Its velocity at $t=2~\mathrm{s}$ is质点的位置为 $x(t) = 3t^2 - 12t + 5$（SI 单位）。其在 $t=2~\mathrm{s}$ 时的速度为
(A) $-12~\mathrm{m/s}$
(B) $0~\mathrm{m/s}$
(C) $6~\mathrm{m/s}$
(D) $12~\mathrm{m/s}$
Answer:答案：(B)
$$v(t) = \frac{dx}{dt} = 6t - 12$$
$$v(2) = 6(2) - 12 = 0~\mathrm{m/s}$$
The particle is momentarily at rest at $t=2~\mathrm{s}$ (a turning point of $x(t)$). Choice (A) is $v(0)$; (D) is $v(4)$.质点在 $t=2~\mathrm{s}$ 时瞬间静止（即 $x(t)$ 的转折点）。选项 (A) 是 $v(0)$，(D) 是 $v(4)$。
Insight.要点。Differentiate the position function before substituting $t=2$; a zero velocity means the particle is instantaneously at rest, not that it was never moving. Check the sign of $v(t)$ on either side to confirm the turning point.先对位置函数求导，再代入 $t=2$；速度为零表示质点瞬时静止，而不是从未运动。检查 $v(t)$ 在两侧的符号以确认转折点。
$a$ is the second derivative of the position polynomial; for a quadratic in $t$, this is a constant.$a$ 是位置多项式的二阶导数；对于关于 $t$ 的二次多项式，该值为常数。
Insight.要点。For a polynomial $x(t)$, acceleration is the second derivative. Because $x$ is quadratic here, $a$ is constant, so the whole motion fits the uniform-acceleration kinematic equations without any approximation.对多项式 $x(t)$，加速度是二阶导数。这里 $x$ 是二次函数，故 $a$ 为常数，整个运动可无近似地使用匀加速运动学方程。
A velocity vs. time graph is described in Q4 of the practice set: $v$ is constant on $[0,2]~\mathrm{s}$, slopes down on $[2,4]~\mathrm{s}$, and is constant again on $[4,6]~\mathrm{s}$. During which interval is the particle's acceleration negative and nonzero?练习题 Q4 描述了一张速度-时间图：$v$ 在 $[0,2]~\mathrm{s}$ 内为常数，在 $[2,4]~\mathrm{s}$ 内下降，在 $[4,6]~\mathrm{s}$ 内再次为常数。在哪段时间区间内质点的加速度为负且不为零？
(A) $0 < t < 2~\mathrm{s}$
(B) $2 < t < 4~\mathrm{s}$
(C) $4 < t < 6~\mathrm{s}$
(D)nowhere on the graph图中任何地方均不满足
Answer:答案：(B)
Acceleration is the slope of the $v$-$t$ graph. The flat segments (A) and (C) have $a = 0$. Only the descending segment on $2 < t < 4$ has $a < 0$ and nonzero.加速度是 $v$-$t$ 图的斜率。平坦段 (A) 和 (C) 的 $a = 0$；只有 $2 < t < 4$ 的下降段满足 $a < 0$ 且不为零。
Insight.要点。Acceleration is the slope of a $v$-$t$ graph, not the value of $v$. A flat segment has $a=0$ even when the object is moving; only a segment that falls has negative, nonzero $a$.加速度是 $v$-$t$ 图的斜率，而不是 $v$ 的数值。即使物体正在运动，平坦段的 $a=0$；只有下降段才具有负且非零的 $a$。
A car starts from rest and accelerates uniformly to $20~\mathrm{m/s}$ over $5~\mathrm{s}$. The distance it travels in this interval is一辆汽车从静止出发，在 $5~\mathrm{s}$ 内匀加速至 $20~\mathrm{m/s}$。该过程中行驶的路程为
(A) $25~\mathrm{m}$
(B) $50~\mathrm{m}$
(C) $75~\mathrm{m}$
(D) $100~\mathrm{m}$
Answer:答案：(B)
Use the average-velocity shortcut for uniform acceleration:利用匀加速运动的平均速度简算：
Insight.要点。The average-velocity shortcut $d=(v_0+v)t/2$ is valid only for uniform acceleration. When $a$ varies with time, distance is the integral of $v(t)$, not the midpoint times $t$.平均速度简算 $d=(v_0+v)t/2$ 只在匀加速时成立。当 $a$ 随时间变化时，路程应是对 $v(t)$ 的积分，而不是中点速度乘 $t$。
Q6MEDIUM1.3 Free Fall1.3 自由落体No Calculator
A ball is dropped from rest from a height $H$ above level ground. Air resistance is negligible. Its speed just before impact is一只球从距水平地面高度 $H$ 处由静止释放，空气阻力不计。其落地前瞬间的速率为
(A) $\sqrt{gH}$
(B) $\sqrt{2gH}$
(C) $2gH$
(D) $gH$
Answer:答案：(B)
Apply $v^2 = v_0^2 + 2g\,\Delta y$ with $v_0 = 0$ and $\Delta y = H$:代入 $v^2 = v_0^2 + 2g\,\Delta y$，其中 $v_0 = 0$，$\Delta y = H$：
$$v^2 = 2gH \quad\Longrightarrow\quad v = \sqrt{2gH}$$
Choices (C) and (D) have wrong units (m²/s² and m²/s², not m/s).选项 (C) 和 (D) 的单位有误（m²/s²，而非 m/s）。
Insight.要点。Choose the equation with the knowns and unknown you actually have: $v_0=0$, displacement $H$, and acceleration $g$ point to $v^2=v_0^2+2gH$. Dimensional analysis also rejects any answer not in $m/s$.选择包含已知量与所求量的方程：$v_0=0$、位移 $H$、加速度 $g$，故用 $v^2=v_0^2+2gH$。量纲分析也可排除任何单位不是 $m/s$ 的答案。
A particle has acceleration $a(t) = 6t~\mathrm{m/s^2}$ and starts from rest at $x=0$. Its position at $t=2~\mathrm{s}$ is一质点的加速度为 $a(t) = 6t~\mathrm{m/s^2}$，从 $x=0$ 处由静止开始运动。其在 $t=2~\mathrm{s}$ 时的位置为
(A) $4~\mathrm{m}$
(B) $8~\mathrm{m}$
(C) $12~\mathrm{m}$
(D) $24~\mathrm{m}$
Answer:答案：(B)
Integrate twice from rest at the origin:从原点由静止开始对加速度积分两次：
$$v(t) = \int_0^t 6\tau\,d\tau = 3t^2$$
$$x(t) = \int_0^t 3\tau^2\,d\tau = t^3$$
$$x(2) = 2^3 = 8~\mathrm{m}$$
Trap (D) is $a(2) \cdot t$, treating $a$ as constant.陷阱 (D) 将 $a$ 视为常数，错误地计算了 $a(2) \cdot t$。
Insight.要点。The constant-acceleration formulas are shortcuts for integrals. Here $a$ depends on $t$, so integrate twice: first $a\to v$, then $v\to x$. Never treat $a(2)$ as if it were the constant acceleration over the interval.匀加速公式本质上是积分的捷径。这里 $a$ 随 $t$ 变化，所以需要积分两次：先由 $a$ 得 $v$，再由 $v$ 得 $x$。不要把 $a(2)$ 当成整个区间的恒定加速度。
Q8MEDIUM1.4 Relative Motion1.4 相对运动No Calculator
A boat heads due north across a river at $4~\mathrm{m/s}$ relative to the water. The river flows due east at $3~\mathrm{m/s}$. The boat's speed relative to the bank is一艘船相对于水以 $4~\mathrm{m/s}$ 向正北方向横渡河流，河流以 $3~\mathrm{m/s}$ 向正东方流动。船相对于河岸的速率为
(A) $1~\mathrm{m/s}$
(B) $5~\mathrm{m/s}$
(C) $7~\mathrm{m/s}$
(D) $\sqrt{7}~\mathrm{m/s}$
Answer:答案：(B)
Velocities add as vectors. The components are perpendicular (north and east), so the resultant speed is速度按矢量相加。两分量相互垂直（北向与东向），合速率为
Trap (C) adds magnitudes directly (only valid for parallel vectors); (A) subtracts them (only for anti-parallel).陷阱 (C) 直接相加大小（仅对平行矢量有效）；(A) 相减（仅对反向平行矢量有效）。
Insight.要点。Perpendicular velocity components combine by the Pythagorean theorem, never by adding $3+4$. Write $\vec v_{\text{bank}}=\vec v_{\text{boat/water}}+\vec v_{\text{water/bank}}$ and only then take the magnitude.相互垂直的速度分量用勾股定理合成，绝不能直接相加 $3+4$。先写 $\vec v_{\text{岸}}=\vec v_{\text{船/水}}+\vec v_{\text{水/岸}}$，再取模。
Q9MEDIUM1.5 Projectile Motion1.5 抛体运动Calculator
A ball is launched from ground level with initial speed $25~\mathrm{m/s}$ at $40^\circ$ above the horizontal. Take $g=9.8~\mathrm{m/s^2}$. Its time of flight (return to launch height) is closest to一只球以 $25~\mathrm{m/s}$ 的初速度从地面以仰角 $40^\circ$ 抛出，取 $g=9.8~\mathrm{m/s^2}$。其飞行时间（回到发射高度所需时间）最接近
Trap (A) uses $v_0\sin\theta/g$ (time to apex, not full flight).陷阱 (A) 使用的是 $v_0\sin\theta/g$（上升到最高点的时间，不是全程时间）。
Insight.要点。For level-ground flight the return-to-launch-height time is twice the time to the apex. The factor $2$ disappears if the projectile lands above or below its starting height; then solve the vertical quadratic.在平地上飞行时，回到发射高度的时间是到达最高点时间的两倍。若落点高于或低于起点，这个因子 $2$ 不再成立，应解竖直方向的二次方程。
A projectile is launched from ground level with speed $v_0$ at angle $\theta$. Neglect air resistance. Its horizontal range is maximized when $\theta$ equals一抛体以速度 $v_0$、仰角 $\theta$ 从地面发射，不计空气阻力。当 $\theta$ 等于多少时水平射程最大？
(A) $30^\circ$
(B) $45^\circ$
(C) $60^\circ$
(D) $90^\circ$
Answer:答案：(B)
Range $R = (v_0^2/g)\sin(2\theta)$ is maximized when $\sin(2\theta)=1$, i.e. $2\theta = 90^\circ$, so $\theta = 45^\circ$. The $30^\circ$ and $60^\circ$ pair give equal ranges below the maximum.射程 $R = (v_0^2/g)\sin(2\theta)$ 在 $\sin(2\theta)=1$ 时取最大值，即 $2\theta = 90^\circ$，故 $\theta = 45^\circ$。$30^\circ$ 和 $60^\circ$ 的射程相等，但均小于最大值。
Insight.要点。Range depends on $\sin(2\theta)$, so $30^\circ$ and $60^\circ$ are mirror angles with equal range, and $45^\circ$ is maximum. This symmetry exists only for level ground and negligible drag.射程取决于 $\sin(2\theta)$，故 $30^\circ$ 与 $60^\circ$ 是一对互为余角、射程相等的角，$45^\circ$ 时最大。这种对称性仅在平地和忽略阻力时成立。
Q11HARD1.2 Average vs. Instantaneous1.2 平均速度与瞬时速度No Calculator
A particle's position is $x(t)=t^3 - 6t^2 + 9t$ (SI). On the interval $0 \le t \le 4~\mathrm{s}$, the particle's average velocity equals its instantaneous velocity at $t=$质点的位置为 $x(t)=t^3 - 6t^2 + 9t$（SI 单位）。在区间 $0 \le t \le 4~\mathrm{s}$ 上，质点的平均速度等于其瞬时速度时，$t=$
Both roots lie in $[0,4]$. Trap (D) drops the factor of 2 from the discriminant.两根均在 $[0,4]$ 内。陷阱 (D) 在判别式中漏掉了因子 2。
Insight.要点。The Mean Value Theorem says the secant slope $\bar v$ is attained by some instantaneous slope on the interval. Set $v(t)=\bar v$ and solve, but discard roots outside $[0,4]$.中值定理表明，割线斜率 $\bar v$ 会在区间内某处被瞬时斜率取到。令 $v(t)=\bar v$ 求解，但应舍去 $[0,4]$ 之外的根。
Q12HARD1.3 Free Fall + Reaction1.3 自由落体与反应Calculator
A stone is thrown straight up from the edge of a $40~\mathrm{m}$ cliff with initial speed $15~\mathrm{m/s}$. Take $g=9.8~\mathrm{m/s^2}$ and ignore air resistance. The stone's speed when it strikes the ground at the cliff's base is closest to一块石头以 $15~\mathrm{m/s}$ 的初速度从 $40~\mathrm{m}$ 高的悬崖边缘竖直向上抛出，取 $g=9.8~\mathrm{m/s^2}$，忽略空气阻力。石头落到崖底时的速率最接近
(A) $19~\mathrm{m/s}$
(B) $23~\mathrm{m/s}$
(C) $32~\mathrm{m/s}$
(D) $43~\mathrm{m/s}$
Answer:答案：(C)
The path detail (up then down) is irrelevant: $v^2 = v_0^2 + 2g\,|\Delta y|$ with $|\Delta y| = 40~\mathrm{m}$ downward overall.路径细节（先升后降）无关紧要，整体向下位移为 $|\Delta y| = 40~\mathrm{m}$：
$$v^2 = (15)^2 + 2(9.8)(40) = 225 + 784 = 1009$$
$$v = \sqrt{1009} \approx 31.8~\mathrm{m/s}$$
Trap (D) is $v_0 + \sqrt{2g\cdot 40}$, which double-counts the initial speed.陷阱 (D) 为 $v_0 + \sqrt{2g\cdot 40}$，重复计入了初速度。
Insight.要点。With constant $g$, $v^2=v_0^2+2g\Delta y$ depends only on the net vertical displacement, not the up-and-down path. Use the signed displacement and never add the initial speed twice.在 $g$ 恒定时，$v^2=v_0^2+2g\Delta y$ 只取决于净竖直位移，与先升后降的路径无关。代入有符号位移，且不要重复计入初速度。
Q13HARD1.5 2-D Motion1.5 二维运动No Calculator
A particle moves in the $xy$-plane with $\vec r(t) = (3t)\,\hat\imath + (4t - t^2)\,\hat\jmath$ (SI). Its speed at $t=1~\mathrm{s}$ is一质点在 $xy$ 平面内运动，其位置矢量为 $\vec r(t) = (3t)\,\hat\imath + (4t - t^2)\,\hat\jmath$（SI 单位）。其在 $t=1~\mathrm{s}$ 时的速率为
Trap (D) is $|\vec r(1)| = \sqrt{9 + 9}$, the magnitude of position, not velocity. Trap (A) keeps only the $y$-component.陷阱 (D) 是 $|\vec r(1)| = \sqrt{9 + 9}$，即位置的大小，而非速度。陷阱 (A) 只保留了 $y$ 分量。
Insight.要点。To get speed from a vector position, differentiate the components first, then take the magnitude: $|\vec v|=|d\vec r/dt|$. Taking $|\vec r|$ first and then differentiating is a different quantity.由位置矢量求速率时，应先对各分量求导，再取模：$|\vec v|=|d\vec r/dt|$。先取 $|\vec r|$ 再求导得到的是另一个量。
A particle's $a$ vs. $t$ graph is a triangle: $a$ rises linearly from $0$ at $t=0$ to $a_0$ at $t=T/2$, then falls linearly back to $0$ at $t=T$. If the particle starts from rest, its speed at $t=T$ is一质点的 $a$-$t$ 图像为三角形：加速度从 $t=0$ 时的 $0$ 线性增大至 $t=T/2$ 时的 $a_0$，再线性减小至 $t=T$ 时的 $0$。若质点从静止出发，其在 $t=T$ 时的速率为
(A) $\dfrac{a_0 T}{4}$
(B) $\dfrac{a_0 T}{2}$
(C) $a_0 T$
(D) $\dfrac{a_0 T^2}{2}$
Answer:答案：(B)
The change in velocity from rest equals the signed area under $a(t)$. For the triangle:从静止开始的速度变化量等于 $a(t)$ 曲线下的有向面积。对于该三角形：
$$\Delta v = \tfrac{1}{2}\,\text{base}\,\cdot\,\text{height} = \tfrac{1}{2}\,T\,a_0 = \frac{a_0 T}{2}$$
Trap (D) has units of position (area-of-$v$-curve), not speed.陷阱 (D) 的单位是位置（$v$ 曲线下面积），而非速度。
Insight.要点。The change in velocity is the signed area under an $a$-$t$ graph, exactly like displacement being the area under a $v$-$t$ graph. For the triangle, use $\tfrac12\,\text{base}\cdot\text{height}$.速度变化量等于 $a$-$t$ 图下的有向面积，正如位移等于 $v$-$t$ 图下的面积。三角形面积用 $\tfrac12\times$ 底 $\times$ 高。
Q15EASY1.4 Relative Motion1.4 相对运动No Calculator
A train moves due east at $20~\mathrm{m/s}$ relative to the ground. A passenger walks toward the front of the train at $1.5~\mathrm{m/s}$ relative to the train. The passenger's velocity relative to the ground is一列火车相对于地面以 $20~\mathrm{m/s}$ 向正东方行驶。一位乘客相对于火车以 $1.5~\mathrm{m/s}$ 向车头方向行走。该乘客相对于地面的速度为
Trap (D) treats the vectors as perpendicular, but the passenger walks along the train, not across it.陷阱 (D) 将两矢量视为垂直，但乘客是沿火车方向行走，而非横跨。
Insight.要点。Frame labels resolve the sign: $v_{P/G}=v_{P/T}+v_{T/G}$. When both velocities point the same way, add; when opposite, subtract. Use Pythagoras only when they are perpendicular.明确参考系符号：$v_{P/G}=v_{P/T}+v_{T/G}$。两速度同向时相加，反向时相减；只有垂直时才用勾股定理。
Q16MEDIUM1.2 Sign Analysis1.2 符号分析No Calculator
A particle moves along the $x$-axis with velocity $v(t) > 0$ that is decreasing in time. Which statement best describes the motion?一质点沿 $x$ 轴运动，其速度 $v(t) > 0$ 且随时间减小。下列哪个说法最能描述该运动？
(A)The particle is moving in the $+x$ direction and speeding up.质点沿 $+x$ 方向运动且速率增大。
(B)The particle is moving in the $+x$ direction and slowing down.质点沿 $+x$ 方向运动且速率减小。
(C)The particle is moving in the $-x$ direction and speeding up.质点沿 $-x$ 方向运动且速率增大。
(D)The particle is at rest.质点静止。
Answer:答案：(B)
$v > 0$ fixes the direction ($+x$). "Decreasing in time" means $|v|$ is decreasing, the particle is slowing down. Speeding up requires $v$ and $a$ to share a sign; here they have opposite signs.$v > 0$ 确定了运动方向（$+x$）。"随时间减小"意味着 $|v|$ 在减小，即质点在减速。加速要求 $v$ 与 $a$ 同号，但此处二者符号相反。
Insight.要点。Positive velocity means moving in $+x$, not speeding up. An object speeds up only when $v$ and $a$ share a sign; here $v>0$ but decreasing means $a<0$, so the object slows.速度为正只表示沿 $+x$ 运动，并不表示加速。只有 $v$ 与 $a$ 同号时才加速；这里 $v>0$ 但正在减小，说明 $a<0$，物体在减速。
From the same height above level ground, ball A is dropped from rest at the same instant ball B is launched horizontally. Air resistance is negligible. Which ball lands first?球 A 从同一高度由静止释放，同时球 B 从同一高度水平抛出，空气阻力不计。哪个球先落地？
(A)Ball A (dropped)球 A（自由落体）
(B)Ball B (horizontal)球 B（水平抛出）
(C)They land simultaneously.两球同时落地。
(D)Depends on ball B's horizontal speed.取决于球 B 的水平速度。
Answer:答案：(C)
Horizontal and vertical motions are independent. Both balls have the same initial height, the same $v_{y0}=0$, and the same vertical acceleration $g$; so $y(t) = y_0 - \tfrac{1}{2}gt^2$ is identical for both. They share a landing time regardless of ball B's horizontal speed.水平运动与竖直运动相互独立。两球初始高度相同，竖直初速度均为 $v_{y0}=0$，竖直加速度均为 $g$，故 $y(t) = y_0 - \tfrac{1}{2}gt^2$ 对两球完全相同。无论球 B 的水平速度如何，两球落地时间相同。
Insight.要点。Landing time is a vertical problem: same starting height, same $v_{y0}=0$, and same $g$ give the same $t$. Horizontal launch speed changes where the ball lands, never when.落地时间属于竖直方向问题：起点高度相同、$v_{y0}=0$ 相同、$g$ 相同，则 $t$ 相同。水平发射速度只改变落点位置，不改变落地时间。
Q18HARD1.5 Projectile + Energy1.5 抛体与能量Calculator
A projectile is launched horizontally from a cliff of height $h$ with initial speed $v_0$. Air resistance is negligible. Its speed at the moment of impact is一抛体以初速度 $v_0$ 从高度为 $h$ 的悬崖顶端水平抛出，空气阻力不计。其落地瞬间的速率为
Or, by energy conservation: $\tfrac{1}{2}mv^2 = \tfrac{1}{2}mv_0^2 + mgh$ gives the same result. Trap (D) adds magnitudes of perpendicular vectors.或用能量守恒：$\tfrac{1}{2}mv^2 = \tfrac{1}{2}mv_0^2 + mgh$，结果相同。陷阱 (D) 将垂直矢量的大小直接相加。
Insight.要点。When only speed and height are involved, energy conservation is often the cleanest route: $\tfrac12 mv^2=\tfrac12 mv_0^2+mgh$. If using components instead, combine $v_x$ and $v_y$ by Pythagoras, not by direct addition.当只涉及速率和高度时，能量守恒往往最简洁：$\tfrac12 mv^2=\tfrac12 mv_0^2+mgh$。若用分量法，则用勾股定理合成 $v_x$ 与 $v_y$，不能直接相加。
Each FRQ below restates the stem, then walks each part with the canonical AP-style structure: set up, execute, evaluate. Numbers use $g = 9.8~\mathrm{m/s^2}$.以下每道自由解答题均重述题干，并按 AP 标准结构逐步展示：建立模型、执行计算、评估结果。数值计算取 $g = 9.8~\mathrm{m/s^2}$。
A particle moves along a straight line with acceleration $a(t) = 6 - 2t~\mathrm{(m/s^2)}$. At $t=0$, the particle is at $x=0$ with velocity $v_0 = 0$.一质点沿直线运动，其加速度为 $a(t) = 6 - 2t~\mathrm{(m/s^2)}$。在 $t=0$ 时，质点位于 $x=0$，初速度 $v_0 = 0$。
At $t = 0$ this is the initial condition. At $t = 6~\mathrm{s}$ check the sign of $a(6) = 6 - 12 = -6 < 0$: $v$ is decreasing through zero, so the particle reverses direction at $t = 6~\mathrm{s}$.$t = 0$ 为初始条件。在 $t = 6~\mathrm{s}$ 时，检验 $a(6) = 6 - 12 = -6 < 0$：$v$ 过零点时递减，故质点在 $t = 6~\mathrm{s}$ 时改变运动方向。
(c)Maximum speed on $0 \le t \le 5~\mathrm{s}$.$0 \le t \le 5~\mathrm{s}$ 内的最大速率。Set $a(t) = 0$: $t = 3~\mathrm{s}$ (interior critical point in the interval).令 $a(t) = 0$：$t = 3~\mathrm{s}$（区间内的极值点）。
$$v(3) = 6(3) - 3^2 = 9~\mathrm{m/s}$$
Endpoints: $v(0) = 0$, $v(5) = 30 - 25 = 5~\mathrm{m/s}$. The maximum is $\boxed{9~\mathrm{m/s}~\text{at}~t = 3~\mathrm{s}}$.端点值：$v(0) = 0$，$v(5) = 30 - 25 = 5~\mathrm{m/s}$。最大值为 $\boxed{9~\mathrm{m/s}，发生于~t = 3~\mathrm{s}}$。
(d)Total distance on $0 \le t \le 5~\mathrm{s}$.$0 \le t \le 5~\mathrm{s}$ 内的总路程。$v(t) = t(6-t) \ge 0$ on $[0, 5]$ (the next zero is at $t = 6$), so no sign change inside the interval. Total distance equals $|x(5) - x(0)|$:$v(t) = t(6-t) \ge 0$ 在 $[0, 5]$ 上成立（下一个零点在 $t = 6$），区间内无符号变化。总路程等于 $|x(5) - x(0)|$：
Total distance $= \tfrac{100}{3}~\mathrm{m}$.总路程 $= \tfrac{100}{3}~\mathrm{m}$。
Insight.要点。Distance and displacement agree only while $v(t)$ keeps one sign. For a general $v(t)$, find its zeros, integrate on each piece, and sum the absolute values; do not integrate $v(t)$ blindly over the whole interval.只有当 $v(t)$ 不变号时，路程才等于位移。对一般 $v(t)$，先求零点，再分段积分并取绝对值相加；不能盲目地在整个区间上直接积分。
Launched from the edge of a cliff of height $h = 30~\mathrm{m}$ with $v_0 = 22~\mathrm{m/s}$ at $\theta = 35^\circ$ above horizontal. Air resistance negligible.从高 $h = 30~\mathrm{m}$ 的悬崖边缘以 $v_0 = 22~\mathrm{m/s}$、仰角 $\theta = 35^\circ$ 发射，空气阻力不计。
(a)Components and kinematic equations (origin at launch, $+y$ up).分量与运动学方程（以发射点为原点，$+y$ 向上）。
Insight.要点。Projectile motion splits into independent axes: horizontal position is $v_x t$, while vertical position solves a quadratic because of $g$. Compute the two separately and recombine at the end.抛体运动按轴独立分解：水平位置为 $v_x t$，竖直位置因 $g$ 需解二次方程。分别计算两轴，最后再合成。
FRQ 3HARD1.4 Relative Motion1.4 相对运动Calculator
River flows east at $u = 2.0~\mathrm{m/s}$; width $L = 80~\mathrm{m}$. Swimmer's speed relative to water is $v = 1.5~\mathrm{m/s}$. Swimmer enters at the south bank.河流以 $u = 2.0~\mathrm{m/s}$ 向东流动，河宽 $L = 80~\mathrm{m}$。游泳者相对于水的速度为 $v = 1.5~\mathrm{m/s}$，从南岸入水。
(a)Aiming due north.正北方向游进。The north (across-stream) velocity is the full $v = 1.5~\mathrm{m/s}$; the east drift is the current $u = 2.0~\mathrm{m/s}$.垂直河流方向（北）的速度为 $v = 1.5~\mathrm{m/s}$；东向漂移由水流 $u = 2.0~\mathrm{m/s}$ 决定。
(b)Can the swimmer reach a point directly across?游泳者能否到达正对岸的点？To cancel the current the swimmer would need an upstream velocity component $v\sin\theta = u$, i.e. $\sin\theta = u/v = 2/1.5 = 1.33$. Impossible, $\sin\theta \le 1$.要抵消水流，游泳者需要 $v\sin\theta = u$，即 $\sin\theta = u/v = 2/1.5 = 1.33$，不可能实现（$\sin\theta \le 1$）。Minimum-drift heading.最小漂移航向。Let $\theta$ be the angle west of north. Ground velocity: east component $u - v\sin\theta$; north component $v\cos\theta$. Time across is $L/(v\cos\theta)$, so设 $\theta$ 为偏西北的角度。相对地面速度：东分量 $u - v\sin\theta$，北分量 $v\cos\theta$。横渡时间为 $L/(v\cos\theta)$，故
Trade-off: the minimum-drift heading costs ~27 s more crossing time than aiming due north (53 s) but cuts drift by ~36 m.权衡：与正北方向（53 s）相比，最小漂移航向多用约 27 s，但漂移减少约 36 m。
Insight.要点。Cancelling a current requires an upstream speed at least as large as the current: $v\sin\theta=u$ needs $u/v\le 1$. When $u>v$, do not search for cancellation; minimize the downstream drift instead.抵消水流要求逆流方向的分速度至少等于水流：$v\sin\theta=u$ 需要 $u/v\le 1$。当 $u>v$ 时，不要试图抵消，而应最小化下游漂移。
FRQ 4HARD1.2 / 1.3 Data + Calculus1.2 / 1.3 数据与微积分Calculator
A cart on a track is released; position recorded with a motion sensor:一辆小车在轨道上释放，运动传感器记录其位置：
$t$ (s)
0.0
0.2
0.4
0.6
0.8
1.0
$x$ (m)
0.00
0.04
0.16
0.36
0.64
1.00
(a)Consistency with constant $a$; numeric value.与匀加速运动的一致性；加速度数值。For constant $a$ from rest at the origin, $x = \tfrac{1}{2}at^2$, so $x$ should scale as $t^2$. The table's $x$ values equal $t^2$ exactly at every sample:从原点由静止开始的匀加速运动满足 $x = \tfrac{1}{2}at^2$，故 $x$ 应与 $t^2$ 成正比。表中每个采样点的 $x$ 值均精确等于 $t^2$：
$$\frac{x}{t^2} = 1.00~\mathrm{m/s^2}~\text{at every point}$$
Prediction from (a): $v(t) = at = (2)(0.5) = 1.0~\mathrm{m/s}$. The centered difference agrees exactly (no surprise, for a polynomial of degree 2 the centered difference is the analytic derivative).(a) 的预测值：$v(t) = at = (2)(0.5) = 1.0~\mathrm{m/s}$。中心差分与解析导数完全吻合（对于二次多项式这并不意外）。
(c)"Doubling the tilt doubles $a$", kinematics or dynamics?"倾角翻倍则加速度翻倍"，运动学还是动力学？Verdict: dynamics is required, and even then the claim is only approximate.结论：需要引入动力学，且即便如此该结论也只是近似成立。Constant-$a$ kinematics ($x = \tfrac{1}{2}at^2$, $v = at$) tells you only that $a$ is constant, it doesn't predict how $a$ varies with the experimental setup. To connect $a$ to the tilt angle you need Newton's second law: on a frictionless incline at angle $\alpha$,匀加速运动学（$x = \tfrac{1}{2}at^2$，$v = at$）只能说明 $a$ 是常数，但无法预测 $a$ 随实验条件如何变化。要将 $a$ 与倾角联系起来，需要牛顿第二定律：在倾角为 $\alpha$ 的无摩擦斜面上，
$$a = g\sin\alpha$$
Doubling $\alpha$ doubles $\sin\alpha$ only in the small-angle limit ($\sin\alpha \approx \alpha$). For larger tilts, $\sin(2\alpha) \ne 2\sin\alpha$, so the claim is generally false even after invoking dynamics.只有在小角近似（$\sin\alpha \approx \alpha$）下，$\alpha$ 翻倍才能使 $\sin\alpha$ 翻倍。对于较大的倾角，$\sin(2\alpha) \ne 2\sin\alpha$，即便引入动力学该结论也普遍不成立。
Insight.要点。Kinematics can test whether $a$ is constant, but it cannot predict how $a$ changes with the apparatus. That step requires a model such as Newton's second law; keep data-fitting and physics modeling distinct.运动学可以检验 $a$ 是否恒定，但不能预测 $a$ 如何随装置变化。这一步需要牛顿第二定律等模型；要把数据拟合与物理建模区分开。