运动学

Kinematics 运动学 describes motion. Two kinds of quantity:

• A scalar 标量 has only size (magnitude 大小): distance, speed, mass, time.
• A vector 矢量 has magnitude and direction: displacement 位移, velocity 速度, acceleration 加速度, force.

In this calculus-based course you resolve 分解 vectors into components 分量 and add them component by component; a vector's magnitude is $\sqrt{v_x^2+v_y^2+\cdots}$. Physics C also writes vectors with unit vectors 单位矢量: $\vec{v}=v_x\hat{i}+v_y\hat{j}$, where $\hat{i}$ and $\hat{j}$ are directions of length one along $x$ and $y$ – handy because calculus can then act on each component separately.

Because motion can change continuously, define the rates as derivatives 导数:

Reversing this, integrate 积分 to recover velocity and position 位置: Distinguish the average from the instantaneous: average velocity 平均速度 is $\Delta x/\Delta t$ over an interval, while instantaneous velocity 瞬时速度 is the derivative at one moment – the slope 斜率 of the tangent line 切线, and what a speedometer 速度计 shows.

For constant acceleration these give the familiar kinematic equations 运动学方程 ($v=v_0+at$, $x=x_0+v_0t+\tfrac12at^2$, $v^2=v_0^2+2a\,\Delta x$). The most important constant-$a$ case is free fall 自由落体: near Earth's surface every object, heavy or light, accelerates downward at $g=9.8\ \text{m/s}^2$ once air resistance 空气阻力 is negligible 可忽略的. When $a$ or $v$ varies with time, use the integrals directly.

Worked example. A particle moves with $x(t)=2t^3-3t^2$ (metres). Differentiate for velocity and acceleration:

At $t=2\ \text{s}$, $v=24-12=12\ \text{m/s}$ and $a=24-6=18\ \text{m/s}^2$. This calculus link – not just the constant-$a$ formulas – is what distinguishes Physics C.

Worked example. A particle starts from rest at the origin with $a(t)=6t$. Integrate: $v=\int 6t\,dt=3t^2$ and $x=\int 3t^2\,dt=t^3$. At $t=2\ \text{s}$, $v=12\ \text{m/s}$ and $x=8\ \text{m}$.

Move fluently between description, graph, table, and equation:

• On a position–time graph, the slope ($dx/dt$) is velocity.
• On a velocity–time graph, the slope is acceleration and the area ($\int v\,dt$) is displacement.
• On an acceleration–time graph, the area ($\int a\,dt$) is the change in velocity.

Slopes are derivatives; areas are integrals – the two are inverse operations 逆运算 here.

All motion is relative to a reference frame 参考系. Combine velocities by vector addition 矢量相加: $\vec{v}_{A/C}=\vec{v}_{A/B}+\vec{v}_{B/C}$. This handles boats crossing rivers and passengers on moving vehicles – relative motion 相对运动 problems. Read the subscripts as a chain: "A relative to B" plus "B relative to C" gives "A relative to C".

Worked example. A boat that moves at $4.0\ \text{m/s}$ in still water heads straight across a river flowing at $3.0\ \text{m/s}$. Relative to the ground the boat moves at $\sqrt{4.0^2+3.0^2}=5.0\ \text{m/s}$, angled downstream. If the river is $80\ \text{m}$ wide, the crossing still takes $t=\dfrac{80}{4.0}=20\ \text{s}$ – only the across-stream component crosses the river; the current just carries the boat $60\ \text{m}$ downstream.

In two or three dimensions, position is a vector $\vec{r}(t)=\langle x(t),y(t),z(t)\rangle$, and velocity and acceleration are its successive derivatives – each component handled independently 独立地.

Worked example. $\vec{r}(t)=\big(2t^2\big)\hat{i}+\big(4t-t^3\big)\hat{j}$ (metres). Differentiating each component: $\vec{v}=4t\,\hat{i}+(4-3t^2)\,\hat{j}$ and $\vec{a}=4\,\hat{i}-6t\,\hat{j}$. At $t=1\ \text{s}$: $\vec{v}=4\hat{i}+1\hat{j}$, so the speed is $\sqrt{17}\approx4.1\ \text{m/s}$ – no new physics, just one derivative per component.

For ideal projectile motion 抛体运动 under uniform gravity with air resistance neglected, horizontal and vertical motions are independent, linked only by time $t$: horizontally $a_x=0$ (constant velocity), vertically $a_y=-g$. The path is a parabola 抛物线. This component method extends to any two-dimensional motion where the accelerations along each axis are known.

Worked example. A ball is launched at $20\ \text{m/s}$, $30^{\circ}$ above the horizontal ($g=9.8\ \text{m/s}^2$). The components are $v_{0x}=20\cos30^{\circ}=17.3\ \text{m/s}$ and $v_{0y}=20\sin30^{\circ}=10\ \text{m/s}$. The vertical motion sets the time: total flight $=\dfrac{2v_{0y}}{g}=\dfrac{20}{9.8}=2.0\ \text{s}$, maximum height $=\dfrac{v_{0y}^2}{2g}=5.1\ \text{m}$, and range $=v_{0x}\times2.0=35\ \text{m}$.

• Resolve every vector into components before adding — never add magnitudes at an angle directly.
• On Physics C you are expected to use calculus: velocity is $\vec v=\tfrac{d\vec r}{dt}$ and acceleration $\vec a=\tfrac{d\vec v}{dt}$; reverse with integration.
• Carry and check units and treat direction with signs (choose a positive axis 坐标轴 and stick to it).
• Use the dot product 点积 for work-type quantities and the cross product 叉积 for torque 力矩 and angular momentum 角动量.
• Sketch the vectors — a diagram catches sign and direction errors the algebra hides.

Common mistakes

• Dropping the constant of integration. $v(t) = \int a\,dt + v_0$ and $x(t) = \int v\,dt + x_0$: the initial conditions belong in the answer.
• Finding average velocity from total distance. Average velocity is displacement over time; distance gives average speed.
• Saying a projectile has zero velocity at the top. Only $v_y$ is zero; $v_x$ is unchanged throughout.
• Using the constant-acceleration equations when $a$ depends on time. Differentiate or integrate instead.
• Adding relative velocities as numbers. $\vec v_{A/G} = \vec v_{A/B} + \vec v_{B/G}$ is a vector sum; draw the triangle.