Kinematics

In 1999 NASA lost the Mars Climate Orbiter, a 125-million-dollar spacecraft, because one team gave thruster data in pound-seconds and another read it as newton-seconds. The probe's calculated path was off by about 170 km, and it burned up in the Martian atmosphere. Kinematics – the careful description of where something is and how that changes – sounds simple, but every later unit in this course is built on getting it exactly right: the right quantities, the right directions, and the right units.

Kinematics 运动学 describes motion without asking what causes it. Two kinds of quantity:

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

A vector is drawn as an arrow whose length is proportional to its magnitude. In one dimension, direction is just a sign: choose a positive direction, and anything pointing the other way is negative. So a velocity of $-3\ \text{m/s}$ means 3 m/s in the negative direction – not "less than zero speed".

In this calculus-based course you resolve 分解 vectors into components 分量 and add them component by component. Physics C writes vectors with unit vectors 单位矢量 $\hat{i}$, $\hat{j}$ and $\hat{k}$, which have length one and point along $x$, $y$ and $z$:

A resultant vector is the sum of the components: $\vec{C}=\vec{A}+\vec{B}=(A_x+B_x)\hat{i}+(A_y+B_y)\hat{j}$. A vector can also be given as a magnitude and a direction; convert with $A_x=A\cos\theta$, $A_y=A\sin\theta$ and $\tan\theta=A_y/A_x$. The position vector of a point is $\vec{r}$, and $\hat{r}$ is the unit vector pointing the same way.

Worked example (adding vectors). $\vec{A}=3\hat{i}+4\hat{j}$ m and $\vec{B}=6\ \text{m}$ at $30^\circ$ below the $+x$ axis. Find $\vec{A}+\vec{B}$. $\vec{B}=6\cos30^\circ\,\hat{i}-6\sin30^\circ\,\hat{j}=5.2\hat{i}-3.0\hat{j}$ m. $\vec{C}=(3+5.2)\hat{i}+(4-3.0)\hat{j}=8.2\hat{i}+1.0\hat{j}$ m. $|\vec{C}|=\sqrt{8.2^2+1.0^2}=8.3$ m at $\tan^{-1}(1.0/8.2)=7^\circ$ above $+x$. Adding the magnitudes ($5+6=11$ m) would be wrong.

In the object model, we ignore an object's size, shape and internal parts and treat it as a single point that carries its mass (and charge). A car, a planet or a ball can all be objects when only their overall motion matters.

Displacement is the change in position, $\Delta x=x-x_0$. It depends only on the start and end points: run once round a 400 m track and your distance is 400 m but your displacement is zero.

Average values use only the start and end of a time interval:

An object is accelerating whenever its velocity changes in magnitude, direction, or both – a car going round a bend at a steady 50 km/h is accelerating.

As the time interval shrinks toward zero, the average value approaches the value at one instant. The instantaneous velocity and acceleration are therefore derivatives 导数:

Reversing this, integrate 积分 to recover velocity and position 位置:

So 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 on an $x$–$t$ graph, and what a speedometer 速度计 shows (as a magnitude).

Worked example (differentiating). 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 $a=24-6=18\ \text{m/s}^2$. The particle is momentarily at rest when $v=0$: $6t(t-1)=0$, so at $t=0$ and $t=1$ s. Between them $v<0$, so it moves backwards. Its average velocity from $t=0$ to $t=2$ s is $\dfrac{x(2)-x(0)}{2}=\dfrac{4-0}{2}=2\ \text{m/s}$ – very different from the instantaneous 12 m/s at the end. This calculus link – not just the constant-$a$ formulas – is what distinguishes Physics C.

Worked example (integrating). A particle starts from rest at the origin with $a(t)=6t$. Integrate: $v=\int_0^t 6t\,dt=3t^2$ and $x=\int_0^t 3t^2\,dt=t^3$. At $t=2\ \text{s}$, $v=12\ \text{m/s}$ and $x=8\ \text{m}$. If instead it had started at $x_0=5$ m with $v_0=-4$ m/s, then $v=3t^2-4$ and $x=5-4t+t^3$: the initial conditions are the constants of integration.

Motion can be shown with motion diagrams, graphs, equations, tables and words; free-response questions ask you to move between them.

• On a position–time graph, the slope of the tangent ($dx/dt$) is the instantaneous velocity.
• On a velocity–time graph, the slope is acceleration and the area under the curve is displacement: $\Delta x=\int_{t_1}^{t_2}v\,dt$. Area below the time axis counts as negative displacement.
• On an acceleration–time graph, the area is the change in velocity: $\Delta v=\int_{t_1}^{t_2}a\,dt$.

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

For constant acceleration, integrating $a$ twice gives the three kinematic equations 运动学方程, which work in any single direction:

Choose the equation that leaves out the quantity you neither know nor need. Use them only when $a$ is constant; if $a$ depends on time, integrate.

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 $g$ once air resistance 空气阻力 is negligible 可忽略的. On the AP exam use $g\approx10\ \text{m/s}^2$ whenever a number is needed (9.8 is also accepted).

Worked example (area under a graph). A cart starts at rest. Its acceleration is $a=2.0\ \text{m/s}^2$ for 4.0 s, then $a=-1.0\ \text{m/s}^2$ for 6.0 s. Area under the $a$–$t$ graph: $\Delta v=2.0(4.0)+(-1.0)(6.0)=8.0-6.0=2.0$ m/s, so the final velocity is 2.0 m/s. For displacement, use the $v$–$t$ graph: $v$ rises from 0 to 8.0 m/s, then falls to 2.0 m/s. Area $=\tfrac12(4.0)(8.0)+\tfrac12(8.0+2.0)(6.0)=16+30=46$ m.

Worked example (free fall). A stone is thrown straight up at 15 m/s from the edge of a cliff 20 m high and lands at the base. Take up as positive, $a=-10\ \text{m/s}^2$, and $\Delta y=-20$ m.

so $t=4.0$ s (the negative root is not physical). Landing speed: $v^2=15^2+2(-10)(-20)=625$, so $v=-25$ m/s – 25 m/s downward.

Worked example (when $a$ is not constant). Students drop a coffee filter and fit its speed to $v=A\left(1-e^{-Bt}\right)$, with $A=2.0$ m/s and $B=5.0\ \text{s}^{-1}$.

1. Acceleration: $a=\dfrac{dv}{dt}=ABe^{-Bt}$. At $t=0$, $a=AB=10\ \text{m/s}^2$: at the start it is in free fall. At $t=0.20$ s, $a=10e^{-1}\approx3.7\ \text{m/s}^2$.
2. As $t\to\infty$, $e^{-Bt}\to0$, so $v\to A=2.0$ m/s: $A$ is the terminal speed, and the acceleration falls to zero.
3. Position (starting at $x=0$): $x=\displaystyle\int_0^t A\left(1-e^{-Bt'}\right)dt'=At-\dfrac{A}{B}\left(1-e^{-Bt}\right)$.

The kinematic equations would give wrong answers here, because $a$ changes. If you only have a $v$–$t$ graph, find $a$ at an instant by drawing the tangent there and taking its slope, using two points far apart on the tangent line.

Measuring $g$ with a straight-line graph. A ball is dropped from rest from several heights $h$, and the fall time $t$ is measured. Since $h=\tfrac12gt^2$, a graph of $h$ against $t^2$ is a straight line through the origin with slope $\tfrac12g$.

1. Plot $h$ (vertical) against $t^2$ (horizontal) and draw the best-fit line. Do not join the dots.
2. Slope from two points on the line, far apart: $\dfrac{1.60-0.40}{0.326-0.082}\approx4.92\ \text{m/s}^2$.
3. $g=2\times\text{slope}\approx9.84\ \text{m/s}^2$.
4. Percent error $=\dfrac{|\text{experimental}-\text{accepted}|}{\text{accepted}}\times100=\dfrac{|9.84-9.81|}{9.81}\times100\approx0.3\%$.

Choose the axes so the expected relation becomes $y=mx$ (or $y=mx+c$); then say which physical quantity the slope stands for. To reduce uncertainty, take several readings at each height and average them, and use a wide range of heights.

All motion is measured relative to a reference frame 参考系 – a point of view with its own origin and axes. The frame you choose decides the direction and magnitude of the velocities you measure: a passenger walking forward at 1 m/s in a train moving at 30 m/s moves at 31 m/s relative to the ground.

Measurements in one frame are converted to another by vector addition 矢量相加 or subtraction:

Read the subscripts as a chain: "A relative to B" plus "B relative to C" gives "A relative to C". Also, $\vec{v}_{B/A}=-\vec{v}_{A/B}$. This handles boats crossing rivers, planes in wind and passengers on moving vehicles – relative motion 相对运动 problems.

An inertial reference frame is one that is not accelerating. Observers in different inertial frames measure different velocities but the same acceleration for an object, because a constant frame velocity disappears when you differentiate. Unless told otherwise, assume your frame is inertial.

Worked example (heading straight across). A boat moves at $5.0\ \text{m/s}$ relative to the water. The river flows at $3.0\ \text{m/s}$ and is 80 m wide. The pilot wants to land directly opposite. The boat must point partly upstream so that the upstream component of its velocity cancels the current: $5.0\sin\theta=3.0$, so $\theta=37^\circ$ upstream of straight across. Its across-river speed is $5.0\cos37^\circ=4.0$ m/s, and the crossing takes $\dfrac{80}{4.0}=20$ s. If instead it pointed straight across, it would cross in $\dfrac{80}{5.0}=16$ s but land $3.0\times16=48$ m downstream.

In two or three dimensions, position is a vector $\vec{r}(t)=x(t)\hat{i}+y(t)\hat{j}+z(t)\hat{k}$, and velocity and acceleration are its successive derivatives. Motion is analysed by splitting it into components and treating each with one-dimensional kinematics independently 独立地:

• the velocity and acceleration may be different in each direction, and need not be constant;
• a change in one direction does not change the motion in a perpendicular direction – a bullet dropped and a bullet fired horizontally from the same height hit the ground at the same time.

Worked example (calculus in 2D). $\vec{r}(t)=\big(2t^2\big)\hat{i}+\big(4t-t^3\big)\hat{j}$ (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, the acceleration is zero horizontally and constant vertically: $a_x=0$ (constant horizontal velocity) and $a_y=-g$. The two motions are linked only by the time $t$, and the path is a parabola 抛物线.

Worked example (launch on level ground). A ball is launched at $20\ \text{m/s}$, $30^{\circ}$ above the horizontal ($g=10\ \text{m/s}^2$). 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 $=\dfrac{2v_{0y}}{g}=\dfrac{20}{10}=2.0\ \text{s}$, maximum height $=\dfrac{v_{0y}^2}{2g}=5.0\ \text{m}$, and range $=v_{0x}\times2.0=35\ \text{m}$.

Worked example (launch from a height). A ball rolls off a table 1.25 m high at 3.0 m/s. How far from the table does it land, and how fast is it moving? Vertical: $1.25=\tfrac12(10)t^2$, so $t=0.50$ s. Horizontal: $x=3.0\times0.50=1.5$ m. At landing $v_x=3.0$ m/s and $v_y=-10\times0.50=-5.0$ m/s, so $v=\sqrt{3.0^2+5.0^2}=5.8$ m/s at $\tan^{-1}(5.0/3.0)=59^\circ$ below the horizontal.

• 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 and include the initial conditions.
• Use $g\approx10\ \text{m/s}^2$ unless told otherwise.
• Carry and check units and treat direction with signs (choose a positive axis 坐标轴 and stick to it).
• Read graphs both ways: slope gives the next derivative, area gives the change in the previous quantity.
• In relative-motion problems, write the subscript chain $\vec v_{A/C}=\vec v_{A/B}+\vec v_{B/C}$ before substituting.
• 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 (later units).
• 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.
• Thinking zero velocity means zero acceleration. At the top of a vertical throw $v=0$ but $a=-g$.