Companion to the IB-Style Practice SetIB 风格练习题的解析配套
Syllabus A1.1 to A1.6考纲 A1.1 至 A1.6PHYSICS HL
A cyclist rides $100\ \mathrm{m}$ east, then $40\ \mathrm{m}$ north, taking $25\ \mathrm{s}$. (a) distance and displacement; (b) average speed and average velocity.骑行者向东 $100\ \mathrm{m}$，再向北 $40\ \mathrm{m}$，用时 $25\ \mathrm{s}$。(a) 路程与位移；(b) 平均速率与平均速度。
Distance is the path length: $d = 100 + 40 = 140\ \mathrm{m}$. (A1)
The two legs are perpendicular, so the displacement magnitude follows from Pythagoras:
$$ |\vec{s}| = \sqrt{100^{2} + 40^{2}} = \sqrt{11600} \approx 108\ \mathrm{m}. $$(A1)
Average speed uses distance: $v_{\text{avg}} = \dfrac{d}{t} = \dfrac{140}{25} = 5.6\ \mathrm{m\,s^{-1}}$. (A1)
Average velocity uses displacement: $|\vec{v}_{\text{avg}}| = \dfrac{|\vec{s}|}{t} = \dfrac{107.7}{25} \approx 4.3\ \mathrm{m\,s^{-1}}$. (A1)
路程是路径长度：$d = 100 + 40 = 140\ \mathrm{m}$。(A1)
两段路径互相垂直，故位移大小由勾股定理给出：
$$ |\vec{s}| = \sqrt{100^{2} + 40^{2}} = \sqrt{11600} \approx 108\ \mathrm{m}. $$(A1)
平均速率用路程：$v_{\text{avg}} = \dfrac{d}{t} = \dfrac{140}{25} = 5.6\ \mathrm{m\,s^{-1}}$。(A1)
平均速度用位移：$|\vec{v}_{\text{avg}}| = \dfrac{|\vec{s}|}{t} = \dfrac{107.7}{25} \approx 4.3\ \mathrm{m\,s^{-1}}$。(A1)
A car at $28\ \mathrm{m\,s^{-1}}$ brakes uniformly to rest over $98\ \mathrm{m}$. (a) choose the suvat; (b) deceleration; (c) stopping time.汽车以 $28\ \mathrm{m\,s^{-1}}$ 均匀制动，经 $98\ \mathrm{m}$ 停下。(a) 选 suvat；(b) 减速度；(c) 停车时间。
The four pieces of data are $u = 28$, $v = 0$, $s = 98$, with $a$ wanted; the time $t$ is neither given nor asked. (A1)
Choose the suvat that omits $t$: $v^{2} = u^{2} + 2as$. It uses exactly the known quantities in one step. (R1)
Take the direction of motion as positive: $0 = 28^{2} + 2a(98)$. (M1)
$$ a = \frac{-28^{2}}{2(98)} = \frac{-784}{196} = -4.0\ \mathrm{m\,s^{-2}}. $$The negative sign shows the acceleration opposes the motion, so the deceleration is $4.0\ \mathrm{m\,s^{-2}}$. (A1)
Use $v = u + at$ with the known $a$: $0 = 28 + (-4.0)t$. (M1)
$$ t = \frac{28}{4.0} = 7.0\ \mathrm{s}. $$(A1)
四个已知量为 $u = 28$、$v = 0$、$s = 98$，待求 $a$；时间 $t$ 既未给也未问。(A1)
选不含 $t$ 的 suvat：$v^{2} = u^{2} + 2as$，恰好一步用上所有已知量。(R1)
取运动方向为正：$0 = 28^{2} + 2a(98)$。(M1)
$$ a = \frac{-28^{2}}{2(98)} = \frac{-784}{196} = -4.0\ \mathrm{m\,s^{-2}}. $$负号表示加速度与运动方向相反，故减速度为 $4.0\ \mathrm{m\,s^{-2}}$。(A1)
用 $v = u + at$ 代入已求 $a$：$0 = 28 + (-4.0)t$。(M1)
$$ t = \frac{28}{4.0} = 7.0\ \mathrm{s}. $$(A1)
Ball thrown up at $21\ \mathrm{m\,s^{-1}}$, up positive, no drag. (a) max height; (b) time to max height; (c) velocity at $t = 3.0\ \mathrm{s}$ and its direction.小球以 $21\ \mathrm{m\,s^{-1}}$ 上抛，向上为正，无阻力。(a) 最大高度；(b) 到最高点时间；(c) $t = 3.0\ \mathrm{s}$ 的速度及方向。
Up is positive, so $a = -g = -9.81\ \mathrm{m\,s^{-2}}$. At the top $v = 0$. Use $v^{2} = u^{2} + 2as$: $0 = 21^{2} + 2(-9.81)H$. (M1)
$$ H = \frac{21^{2}}{2(9.81)} = \frac{441}{19.62} \approx 22.5\ \mathrm{m}. $$(A1)
Use $v = u + at$ with $v = 0$: $0 = 21 + (-9.81)t$. (M1)
$$ t = \frac{21}{9.81} \approx 2.14\ \mathrm{s}. $$(A1)
$v = u + at = 21 + (-9.81)(3.0) = 21 - 29.4 = -8.43\ \mathrm{m\,s^{-1}}$. (M1)
The result is $\approx -8.4\ \mathrm{m\,s^{-1}}$. The negative sign means the ball is moving in the negative (downward) direction at this instant: it has already passed the top and is falling back. (A1)
向上为正，故 $a = -g = -9.81\ \mathrm{m\,s^{-2}}$。最高点 $v = 0$。用 $v^{2} = u^{2} + 2as$：$0 = 21^{2} + 2(-9.81)H$。(M1)
$$ H = \frac{21^{2}}{2(9.81)} = \frac{441}{19.62} \approx 22.5\ \mathrm{m}. $$(A1)
用 $v = u + at$，$v = 0$：$0 = 21 + (-9.81)t$。(M1)
$$ t = \frac{21}{9.81} \approx 2.14\ \mathrm{s}. $$(A1)
$v = u + at = 21 + (-9.81)(3.0) = 21 - 29.4 = -8.43\ \mathrm{m\,s^{-1}}$。(M1)
结果约为 $-8.4\ \mathrm{m\,s^{-1}}$。负号表示此刻小球沿负方向（向下）运动：它已越过最高点并正在回落。(A1)
Train: $0$ to $20\ \mathrm{m\,s^{-1}}$ in $8.0\ \mathrm{s}$, constant for $6.0\ \mathrm{s}$, then to rest in $4.0\ \mathrm{s}$. (a) first acceleration; (b) total distance from the area; (c) meaning of gradient and area.火车：$8.0\ \mathrm{s}$ 内从 $0$ 到 $20\ \mathrm{m\,s^{-1}}$，匀速 $6.0\ \mathrm{s}$，再 $4.0\ \mathrm{s}$ 减速到静止。(a) 首段加速度；(b) 由面积求总距离；(c) 斜率与面积的含义。
Acceleration is the gradient of the velocity-time line: $a = \dfrac{\Delta v}{\Delta t}$. (M1)
$$ a = \frac{20 - 0}{8.0} = 2.5\ \mathrm{m\,s^{-2}}. $$(A1)
The distance is the area under the graph, split into a triangle, a rectangle, and a triangle. (M1)
$$ s = \underbrace{\tfrac{1}{2}(8.0)(20)}_{80} + \underbrace{(6.0)(20)}_{120} + \underbrace{\tfrac{1}{2}(4.0)(20)}_{40}. $$(M1 for setting up all three areas)
$$ s = 80 + 120 + 40 = 240\ \mathrm{m}. $$(A1)
On a velocity-time graph the gradient gives the acceleration and the area beneath the line gives the displacement. (B1)
加速度是速度-时间图线的斜率：$a = \dfrac{\Delta v}{\Delta t}$。(M1)
$$ a = \frac{20 - 0}{8.0} = 2.5\ \mathrm{m\,s^{-2}}. $$(A1)
距离是图线下的面积，分为三角形、矩形、三角形三块。(M1)
$$ s = \underbrace{\tfrac{1}{2}(8.0)(20)}_{80} + \underbrace{(6.0)(20)}_{120} + \underbrace{\tfrac{1}{2}(4.0)(20)}_{40}. $$(列出全部三块面积得 M1)
$$ s = 80 + 120 + 40 = 240\ \mathrm{m}. $$(A1)
在速度-时间图上，斜率给出加速度，图线下的面积给出位移。(B1)
Hailstone $0.080\ \mathrm{kg}$ falls from rest; drag $F_{D} = bv^{2}$, $b = 5.0\times 10^{-3}\ \mathrm{N\,s^{2}\,m^{-2}}$. (a) why a terminal velocity exists; (b) calculate it; (c) sketch the $v$-$t$ graph and say why suvat fails.冰雹 $0.080\ \mathrm{kg}$ 从静止下落；阻力 $F_{D} = bv^{2}$，$b = 5.0\times 10^{-3}\ \mathrm{N\,s^{2}\,m^{-2}}$。(a) 为何存在收尾速度；(b) 计算之；(c) 画 $v$-$t$ 图并说明 suvat 为何失效。
At release the only force is the weight, so the stone accelerates and its speed rises. As $v$ increases, the drag $bv^{2}$ (acting upward, opposite to motion) grows. (M1)
Eventually the drag equals the weight, the resultant force is zero, the acceleration falls to zero, and the speed stops increasing: that constant speed is the terminal velocity. (A1)
At terminal velocity the forces balance: $mg = b\,v_{T}^{2}$. (M1)
$$ v_{T} = \sqrt{\frac{mg}{b}} = \sqrt{\frac{(0.080)(9.81)}{5.0\times 10^{-3}}}. $$(M1 for substitution)
$$ v_{T} = \sqrt{\frac{0.7848}{0.0050}} = \sqrt{156.96} \approx 12.5\ \mathrm{m\,s^{-1}}. $$(A1)
The graph starts at the origin with the steepest slope (initial acceleration $g$, since drag is zero at $v = 0$) and curves over with steadily decreasing slope, approaching the horizontal asymptote $v = v_{T} \approx 12.5\ \mathrm{m\,s^{-1}}$ without crossing it. (M1 shape, A1 asymptote labelled)
The suvat equations require a constant acceleration. Here the acceleration $a = g - \tfrac{b}{m}v^{2}$ changes continuously as $v$ changes, so suvat cannot be applied. (R1)
释放瞬间只有重力，故石子加速、速率上升。随 $v$ 增大，阻力 $bv^{2}$（向上，与运动反向）增大。(M1)
最终阻力等于重力，合力为零，加速度降为零，速率不再增加：这一恒定速率即收尾速度。(A1)
收尾时受力平衡：$mg = b\,v_{T}^{2}$。(M1)
$$ v_{T} = \sqrt{\frac{mg}{b}} = \sqrt{\frac{(0.080)(9.81)}{5.0\times 10^{-3}}}. $$(代入得 M1)
$$ v_{T} = \sqrt{\frac{0.7848}{0.0050}} = \sqrt{156.96} \approx 12.5\ \mathrm{m\,s^{-1}}. $$(A1)
图线从原点出发，初始斜率最陡（初加速度为 $g$，因 $v = 0$ 时阻力为零），随后斜率持续减小、向水平渐近线 $v = v_{T} \approx 12.5\ \mathrm{m\,s^{-1}}$ 趋近但不穿越。(形状 M1，标注渐近线 A1)
suvat 方程要求加速度恒定。这里加速度 $a = g - \tfrac{b}{m}v^{2}$ 随 $v$ 连续变化，故 suvat 不适用。(R1)
Trolley from rest down a ramp; $s$ vs $t^{2}$ data given. (a) show $s$ vs $t^{2}$ is a straight line through the origin and state the gradient; (b) gradient and acceleration; (c) percentage uncertainty in $s$ at $t^{2} = 16.0$; (d) reason the line might miss the origin.小车从静止沿斜面下滑；给出 $s$ 对 $t^{2}$ 的数据。(a) 证明 $s$ 对 $t^{2}$ 为过原点直线并说明斜率；(b) 斜率与加速度；(c) $t^{2} = 16.0$ 处 $s$ 的百分比不确定度；(d) 直线可能不过原点的原因。
The trolley starts from rest, so $u = 0$. From $s = ut + \tfrac{1}{2}at^{2}$ with $u = 0$: (M1)
$$ s = \tfrac{1}{2}a\,t^{2}. $$This has the form $s = (\text{gradient})\times t^{2}$ with no intercept, so a plot of $s$ against $t^{2}$ is a straight line through the origin. (A1)
Comparing with $y = mx$, the gradient is $\tfrac{1}{2}a$. (A1)
Read the gradient from two well-separated points, e.g. $(1.0,\,0.80)$ and $(16.0,\,12.80)$: (M1)
$$ \text{gradient} = \frac{12.80 - 0.80}{16.0 - 1.0} = \frac{12.00}{15.0} = 0.80\ \mathrm{m\,s^{-2}}. $$(A1)
Since the gradient equals $\tfrac{1}{2}a$: $a = 2(0.80) = 1.6\ \mathrm{m\,s^{-2}}$. (A1)
At that point $s = 12.80\ \mathrm{m}$ with absolute uncertainty $\pm 0.10\ \mathrm{m}$: (M1)
$$ \frac{0.10}{12.80}\times 100\% \approx 0.78\% \approx 0.8\%. $$(A1)
A small systematic effect can shift the whole line. For example a zero error in measuring $s$ (the metre rule not aligned with the release point), or the trolley already moving when timing began. (B1)
Such an effect adds a constant to every reading, producing a non-zero intercept, which signals a systematic rather than random error. (R1)
小车从静止出发，故 $u = 0$。由 $s = ut + \tfrac{1}{2}at^{2}$ 且 $u = 0$：(M1)
$$ s = \tfrac{1}{2}a\,t^{2}. $$此式形如 $s = (\text{斜率})\times t^{2}$，无截距，故 $s$ 对 $t^{2}$ 作图为过原点的直线。(A1)
与 $y = mx$ 比较，斜率为 $\tfrac{1}{2}a$。(A1)
用相距较远的两点读斜率，如 $(1.0,\,0.80)$ 与 $(16.0,\,12.80)$：(M1)
$$ \text{斜率} = \frac{12.80 - 0.80}{16.0 - 1.0} = \frac{12.00}{15.0} = 0.80\ \mathrm{m\,s^{-2}}. $$(A1)
因斜率等于 $\tfrac{1}{2}a$：$a = 2(0.80) = 1.6\ \mathrm{m\,s^{-2}}$。(A1)
该点 $s = 12.80\ \mathrm{m}$，绝对不确定度 $\pm 0.10\ \mathrm{m}$：(M1)
$$ \frac{0.10}{12.80}\times 100\% \approx 0.78\% \approx 0.8\%. $$(A1)
微小的系统效应会使整条直线平移。例如测 $s$ 时存在零点误差（米尺未对准释放点），或计时开始时小车已在运动。(B1)
这类效应给每个读数加上一个常量，产生非零截距，表明是系统误差而非随机误差。(R1)
Ball projected horizontally at $18\ \mathrm{m\,s^{-1}}$ from a $45\ \mathrm{m}$ cliff, lands on level ground, no drag. (a) why components are independent; (b) time of flight; (c) horizontal distance; (d) landing speed; (e) angle below horizontal at impact.小球以 $18\ \mathrm{m\,s^{-1}}$ 从 $45\ \mathrm{m}$ 崖顶水平抛出，落到等高地面，无阻力。(a) 为何分量独立；(b) 飞行时间；(c) 水平距离；(d) 落地速率；(e) 落地时与水平方向的夹角。
Gravity acts only vertically, so it changes the vertical velocity but never the horizontal velocity. The horizontal and vertical motions therefore proceed independently and share only the time. (R1)
Vertical motion, taking down as positive: $u_{y} = 0$, $a = g$, $s_{y} = 45$. Use $s_{y} = u_{y}t + \tfrac{1}{2}gt^{2}$: $45 = \tfrac{1}{2}(9.81)t^{2}$. (M1·M1)
$$ t = \sqrt{\frac{2(45)}{9.81}} = \sqrt{9.174} \approx 3.03\ \mathrm{s}. $$(A1)
Horizontally there is no acceleration, so $R = u_{x}\,t = 18 \times 3.03$. (M1)
$$ R \approx 54.5\ \mathrm{m}. $$(A1)
The vertical speed at landing is $v_{y} = u_{y} + gt = 0 + 9.81(3.03) \approx 29.7\ \mathrm{m\,s^{-1}}$. (M1)
The horizontal speed is unchanged, $v_{x} = 18\ \mathrm{m\,s^{-1}}$. Combine the perpendicular components: (M1)
$$ v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{18^{2} + 29.7^{2}} = \sqrt{324 + 882.4} \approx 34.7\ \mathrm{m\,s^{-1}}. $$(A1)
The velocity vector points below the horizontal by an angle $\theta$ with $\tan\theta = \dfrac{v_{y}}{v_{x}}$. (M1)
$$ \theta = \tan^{-1}\!\left(\frac{29.7}{18}\right) = \tan^{-1}(1.65) \approx 58.8^{\circ}. $$(A1·A1)
重力只作用于竖直方向，故它改变竖直速度，却从不改变水平速度。因此水平与竖直运动各自独立进行，只共享时间。(R1)
竖直运动取向下为正：$u_{y} = 0$、$a = g$、$s_{y} = 45$。用 $s_{y} = u_{y}t + \tfrac{1}{2}gt^{2}$：$45 = \tfrac{1}{2}(9.81)t^{2}$。(M1·M1)
$$ t = \sqrt{\frac{2(45)}{9.81}} = \sqrt{9.174} \approx 3.03\ \mathrm{s}. $$(A1)
水平方向无加速度，故 $R = u_{x}\,t = 18 \times 3.03$。(M1)
$$ R \approx 54.5\ \mathrm{m}. $$(A1)
落地时竖直速率 $v_{y} = u_{y} + gt = 0 + 9.81(3.03) \approx 29.7\ \mathrm{m\,s^{-1}}$。(M1)
水平速率不变，$v_{x} = 18\ \mathrm{m\,s^{-1}}$。合成两个垂直分量：(M1)
$$ v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{18^{2} + 29.7^{2}} = \sqrt{324 + 882.4} \approx 34.7\ \mathrm{m\,s^{-1}}. $$(A1)
速度矢量指向水平面以下，夹角 $\theta$ 满足 $\tan\theta = \dfrac{v_{y}}{v_{x}}$。(M1)
$$ \theta = \tan^{-1}\!\left(\frac{29.7}{18}\right) = \tan^{-1}(1.65) \approx 58.8^{\circ}. $$(A1·A1)
Projectile from ground at $30\ \mathrm{m\,s^{-1}}$, $40^{\circ}$ above horizontal, no drag. (a) velocity components; (b) time of flight; (c) range and max height; (d) speed at $t = 1.5\ \mathrm{s}$; (e) angle for maximum range.弹丸自地面以 $30\ \mathrm{m\,s^{-1}}$、与水平夹角 $40^{\circ}$ 发射，无阻力。(a) 速度分量；(b) 飞行时间；(c) 射程与最大高度；(d) $t = 1.5\ \mathrm{s}$ 的速率；(e) 最大射程的发射角。
Resolve the launch velocity: $v_{0x} = v_{0}\cos\theta$, $v_{0y} = v_{0}\sin\theta$. (M1)
$$ v_{0x} = 30\cos 40^{\circ} \approx 23.0\ \mathrm{m\,s^{-1}}, \qquad v_{0y} = 30\sin 40^{\circ} \approx 19.3\ \mathrm{m\,s^{-1}}. $$(A1)
The flight is symmetric about the apex; total time follows from the vertical motion returning to launch height, $T = \dfrac{2v_{0y}}{g}$. (M1·M1)
$$ T = \frac{2(19.28)}{9.81} \approx 3.93\ \mathrm{s}. $$(A1)
Range is the horizontal speed times the time of flight: $R = v_{0x}\,T = 22.98 \times 3.93 \approx 90.3\ \mathrm{m}$. (M1·A1)
Maximum height uses $H = \dfrac{v_{0y}^{2}}{2g} = \dfrac{19.28^{2}}{2(9.81)} \approx 19.0\ \mathrm{m}$. (A1)
Horizontal speed is constant, $v_{x} = 22.98\ \mathrm{m\,s^{-1}}$. Vertical: $v_{y} = v_{0y} - gt = 19.28 - 9.81(1.5) \approx 4.57\ \mathrm{m\,s^{-1}}$. (M1)
$$ v = \sqrt{22.98^{2} + 4.57^{2}} \approx 23.4\ \mathrm{m\,s^{-1}}. $$(A1)
From $R = \dfrac{v_{0}^{2}\sin 2\theta}{g}$, the range is greatest when $\sin 2\theta = 1$, i.e. $2\theta = 90^{\circ}$, so $\theta = 45^{\circ}$. (A1·R1)
分解发射速度：$v_{0x} = v_{0}\cos\theta$、$v_{0y} = v_{0}\sin\theta$。(M1)
$$ v_{0x} = 30\cos 40^{\circ} \approx 23.0\ \mathrm{m\,s^{-1}}, \qquad v_{0y} = 30\sin 40^{\circ} \approx 19.3\ \mathrm{m\,s^{-1}}. $$(A1)
飞行关于最高点对称；总时间由竖直运动回到发射高度给出，$T = \dfrac{2v_{0y}}{g}$。(M1·M1)
$$ T = \frac{2(19.28)}{9.81} \approx 3.93\ \mathrm{s}. $$(A1)
射程为水平速率乘飞行时间：$R = v_{0x}\,T = 22.98 \times 3.93 \approx 90.3\ \mathrm{m}$。(M1·A1)
最大高度用 $H = \dfrac{v_{0y}^{2}}{2g} = \dfrac{19.28^{2}}{2(9.81)} \approx 19.0\ \mathrm{m}$。(A1)
水平速率恒定，$v_{x} = 22.98\ \mathrm{m\,s^{-1}}$。竖直：$v_{y} = v_{0y} - gt = 19.28 - 9.81(1.5) \approx 4.57\ \mathrm{m\,s^{-1}}$。(M1)
$$ v = \sqrt{22.98^{2} + 4.57^{2}} \approx 23.4\ \mathrm{m\,s^{-1}}. $$(A1)
由 $R = \dfrac{v_{0}^{2}\sin 2\theta}{g}$，当 $\sin 2\theta = 1$（即 $2\theta = 90^{\circ}$）时射程最大，故 $\theta = 45^{\circ}$。(A1·R1)
River $80\ \mathrm{m}$ wide flows east at $2.5\ \mathrm{m\,s^{-1}}$; boat $4.0\ \mathrm{m\,s^{-1}}$ relative to water. (a) time across when steered due north; (b) downstream drift and resultant speed; (c) heading to land directly opposite; (d) crossing time on that route and why it is longer.河宽 $80\ \mathrm{m}$，向东流 $2.5\ \mathrm{m\,s^{-1}}$；船相对水 $4.0\ \mathrm{m\,s^{-1}}$。(a) 船头朝正北时横渡时间；(b) 下游漂移与合速度；(c) 到达正对岸所需船头方向；(d) 该航线的横渡时间及为何更长。
Steered straight across, the full boat speed $4.0\ \mathrm{m\,s^{-1}}$ is the across-stream component; the current adds nothing to the crossing direction. (M1)
$$ t = \frac{\text{width}}{v_{\text{across}}} = \frac{80}{4.0} = 20\ \mathrm{s}. $$(A1)
During those $20\ \mathrm{s}$ the current carries the boat downstream: drift $= v_{\text{river}}\times t = 2.5 \times 20 = 50\ \mathrm{m}$. (M1·A1)
The velocity relative to the bank is the vector sum of two perpendicular parts, $4.0$ across and $2.5$ downstream: (M1)
$$ v = \sqrt{4.0^{2} + 2.5^{2}} = \sqrt{16 + 6.25} = \sqrt{22.25} \approx 4.7\ \mathrm{m\,s^{-1}}. $$(A1)
To cancel the drift the upstream component of the boat velocity must equal the current: $4.0\sin\phi = 2.5$, where $\phi$ is measured from the straight-across (north) line. (M1)
$$ \phi = \sin^{-1}\!\left(\frac{2.5}{4.0}\right) = \sin^{-1}(0.625) \approx 39^{\circ}\ \text{upstream of north}. $$(A1)
Now only the across-stream component carries the boat over: $v_{\text{across}} = 4.0\cos\phi = \sqrt{4.0^{2} - 2.5^{2}} = \sqrt{9.75} \approx 3.12\ \mathrm{m\,s^{-1}}$. So $t = 80 / 3.12 \approx 25.6\ \mathrm{s}$. (M1)
This exceeds the $20\ \mathrm{s}$ of part (a) because part of the boat speed is now spent fighting the current upstream, leaving a smaller across-stream component. (A1)
笔直横渡时，整个船速 $4.0\ \mathrm{m\,s^{-1}}$ 都是横向分量；水流对横渡方向无贡献。(M1)
$$ t = \frac{\text{河宽}}{v_{\text{横}}} = \frac{80}{4.0} = 20\ \mathrm{s}. $$(A1)
这 $20\ \mathrm{s}$ 内水流把船带向下游：漂移 $= v_{\text{河}}\times t = 2.5 \times 20 = 50\ \mathrm{m}$。(M1·A1)
相对河岸的速度是两个垂直分量的矢量和，横向 $4.0$、下游 $2.5$：(M1)
$$ v = \sqrt{4.0^{2} + 2.5^{2}} = \sqrt{16 + 6.25} = \sqrt{22.25} \approx 4.7\ \mathrm{m\,s^{-1}}. $$(A1)
要抵消漂移，船速的上游分量须等于水流：$4.0\sin\phi = 2.5$，$\phi$ 从笔直横渡（正北）方向量起。(M1)
$$ \phi = \sin^{-1}\!\left(\frac{2.5}{4.0}\right) = \sin^{-1}(0.625) \approx 39^{\circ}\ \text{偏向上游}. $$(A1)
此时只有横向分量带船过河：$v_{\text{横}} = 4.0\cos\phi = \sqrt{4.0^{2} - 2.5^{2}} = \sqrt{9.75} \approx 3.12\ \mathrm{m\,s^{-1}}$。故 $t = 80 / 3.12 \approx 25.6\ \mathrm{s}$。(M1)
它超过 (a) 的 $20\ \mathrm{s}$，因为现在部分船速用于逆流抵消水流，剩下的横向分量更小。(A1)
Skydiver $75\ \mathrm{kg}$, linear drag $F_{D} = kv$, $k = 15\ \mathrm{N\,s\,m^{-1}}$, from rest. (a) Newton's second law and acceleration at release; (b) terminal velocity; (c) acceleration at $v = 20\ \mathrm{m\,s^{-1}}$; (d) how acceleration changes towards $v_{T}$.跳伞者 $75\ \mathrm{kg}$，线性阻力 $F_{D} = kv$，$k = 15\ \mathrm{N\,s\,m^{-1}}$，从静止下落。(a) 牛顿第二定律与释放时加速度；(b) 收尾速度；(c) $v = 20\ \mathrm{m\,s^{-1}}$ 时加速度；(d) 加速度趋向 $v_{T}$ 时如何变化。
Taking downward as positive, the weight acts down and the drag $kv$ acts up: (M1)
$$ ma = mg - kv. $$At release $v = 0$, so drag is zero and $a_{0} = g = 9.81\ \mathrm{m\,s^{-2}}$. (A1)
At terminal velocity $a = 0$, so $mg = kv_{T}$. (M1)
$$ v_{T} = \frac{mg}{k} = \frac{(75)(9.81)}{15} = \frac{735.75}{15} \approx 49\ \mathrm{m\,s^{-1}}. $$(A1)
Rearrange the equation of motion: $a = g - \dfrac{k}{m}v = 9.81 - \dfrac{15}{75}(20)$. (M1)
$$ a = 9.81 - (0.20)(20) = 9.81 - 4.0 \approx 5.8\ \mathrm{m\,s^{-2}}. $$(A1)
As the speed rises from zero towards $v_{T}$, the acceleration decreases steadily from $g$ down to zero. (A1)
This is because the upward drag $kv$ grows with speed, shrinking the resultant force $mg - kv$ until it vanishes at $v = v_{T}$. (R1)
取向下为正，重力向下、阻力 $kv$ 向上：(M1)
$$ ma = mg - kv. $$释放时 $v = 0$，阻力为零，故 $a_{0} = g = 9.81\ \mathrm{m\,s^{-2}}$。(A1)
收尾时 $a = 0$，故 $mg = kv_{T}$。(M1)
$$ v_{T} = \frac{mg}{k} = \frac{(75)(9.81)}{15} = \frac{735.75}{15} \approx 49\ \mathrm{m\,s^{-1}}. $$(A1)
改写运动方程：$a = g - \dfrac{k}{m}v = 9.81 - \dfrac{15}{75}(20)$。(M1)
$$ a = 9.81 - (0.20)(20) = 9.81 - 4.0 \approx 5.8\ \mathrm{m\,s^{-2}}. $$(A1)
当速率从零升向 $v_{T}$ 时，加速度从 $g$ 稳步减小到零。(A1)
这是因为向上的阻力 $kv$ 随速率增大，使合力 $mg - kv$ 不断减小，直到 $v = v_{T}$ 时为零。(R1)