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Kinematics

AP Physics C: Mechanics · Topic 1

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Handout Vocabulary
1.1

Scalars and Vectors
Syllabus
Learning ObjectiveEssential Knowledge

1.1.A
Describe a scalar or vector quantity using magnitude and direction, as appropriate.

  • 1.1.A.1 Scalars are quantities described by magnitude only; vectors are quantities described by both magnitude and direction.
  • 1.1.A.2 Vectors can be visually modeled as arrows with appropriate direction and lengths proportional to their magnitude.
  • 1.1.A.3 Distance and speed are examples of scalar quantities, while position, displacement, velocity, and acceleration are examples of vector quantities.
  • 1.1.A.4 Vectors can be expressed in unit vector notation or as a magnitude and a direction.
    • 1.1.A.4.i Unit vector notation can be used to represent vectors as the sum of their constituent components in the $x$-, $y$-, and $z$-directions, denoted by $\hat{i}$, $\hat{j}$, and $\hat{k}$, respectively.
      • Equation: $\vec{r} = \left( A\hat{i} + B\hat{j} + C\hat{k} \right)$
    • 1.1.A.4.ii The position vector of a point is given by $\vec{r}$, and the unit vector in the direction of the position vector is denoted $\hat{r}$.
    • 1.1.A.4.iii A resultant vector is the vector sum of the addend vectors' components.
      • Equation: $\vec{C} = \vec{A} + \vec{B}$
      • Equation: $\vec{C} = \left( A_x + B_x \right)\hat{i} + \left( A_y + B_y \right)\hat{j}$
  • 1.1.A.5 In a given one-dimensional coordinate system, opposite directions are denoted by opposite signs.

Source: College Board AP Course and Exam Description

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, velocity, acceleration, force.

In this calculus-based course you resolve 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.

Resolving a vector into its x and y components Resolving a vector into its x and y components

Explore

Explore the dot product

Drag the components of $\vec{A}$ and $\vec{B}$. The dot product $\vec{A}\cdot\vec{B}=A_xB_x+A_yB_y$ reaches zero exactly when the two vectors are perpendicular — the geometry behind $\cos\theta$.

Vocabulary Train
English Chinese Pinyin
Kinematics 运动学 yùn dòng xué
scalar 标量 biāo liàng
magnitude 大小 dà xiǎo
vector 矢量 shǐ liàng
components 分量 fèn liàng
unit vectors 单位矢量 dān wèi shǐ liàng
Exercise sheet
1.2

Displacement, Velocity, and Acceleration
Syllabus
Learning ObjectiveEssential Knowledge

1.2.A
Describe a change in an object's position.

  • 1.2.A.1 When using the object model, the size, shape, and internal configuration are ignored. The object may be treated as a single point with extensive properties such as mass and charge.
  • 1.2.A.2 Displacement is the change in an object's position.
    • Equation: $\Delta x = x - x_0$

1.2.B
Describe the average velocity and acceleration of an object.

  • 1.2.B.1 Averages of velocity and acceleration are calculated considering the initial and final states of an object over an interval of time.
  • 1.2.B.2 Average velocity is the displacement of an object divided by the interval of time in which that displacement occurs.
    • Equation: $\vec{v}_{\text{avg}} = \dfrac{\Delta \vec{x}}{\Delta t}$
  • 1.2.B.3 Average acceleration is the change in velocity divided by the interval of time in which that change in velocity occurs.
    • Equation: $\vec{a}_{\text{avg}} = \dfrac{\Delta \vec{v}}{\Delta t}$
  • 1.2.B.4 An object is accelerating if either the magnitude and/or direction of the object's velocity are changing.
  • 1.2.B.5 Calculating average velocity or average acceleration over a very small time interval yields a value that is very close to the instantaneous velocity or instantaneous acceleration.

1.2.C
Describe the instantaneous position, velocity, and acceleration of an object as a function of time.

  • 1.2.C.1 As the time interval used to calculate the average value of a quantity approaches zero, the average value of that quantity approaches the value of the quantity at that instant, called the instantaneous value.
    • 1.2.C.1.i Instantaneous velocity is the rate of change of the object's position, which is equal to the derivative of position with respect to time.
      • Equation: $\vec{v} = \dfrac{d\vec{r}}{dt}$
      • Equation: $v_x = \dfrac{dx}{dt}$
    • 1.2.C.1.ii Instantaneous acceleration is the rate of change of the object's velocity, which is equal to the derivative of velocity with respect to time.
      • Equation: $\vec{a} = \dfrac{d\vec{v}}{dt}$
      • Equation: $a_x = \dfrac{dv_x}{dt}$
  • 1.2.C.2 Time-dependent functions and instantaneous values of position, velocity, and acceleration can be determined using differentiation and integration.

Source: College Board AP Course and Exam Description

Projectile motion is two independent motions

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

$$v=\frac{dx}{dt},\qquad a=\frac{dv}{dt}=\frac{d^2x}{dt^2}.$$
Reversing this, integrate to recover velocity and position:
$$v(t)=v_0+\int_0^t a\,dt,\qquad x(t)=x_0+\int_0^t v\,dt.$$
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 of the tangent line, and what a speedometer shows.

For constant acceleration these give the familiar kinematic equations ($v=v_0+at$, $x=x_0+v_0t+\tfrac12at^2$, $v^2=v_0^2+2a\,\Delta x$). 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 is negligible. 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:

$$v=\frac{dx}{dt}=6t^2-6t,\qquad a=\frac{dv}{dt}=12t-6.$$
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 $x=8\ \text{m}$.

On a velocity-time graph the gradient is the acceleration and the area is the displacement On a velocity-time graph the gradient is the acceleration and the area is the displacement

Vocabulary Train
English Chinese Pinyin
derivatives 导数 dǎo shù
average velocity 平均速度 píng jūn sù dù
instantaneous velocity 瞬时速度 shùn shí sù dù
free fall 自由落体 zì yóu luò tǐ
Exercise sheet
1.3

Representing Motion
Syllabus
Learning ObjectiveEssential Knowledge

1.3.A
Describe the position, velocity, and acceleration of an object using representations of that object's motion.

  • 1.3.A.1 Motion can be represented by motion diagrams, figures, graphs, equations, and narrative descriptions.
  • 1.3.A.2 For constant acceleration, three kinematic equations can be used to describe instantaneous linear motion in one dimension:
    • Equation: $v_x = v_{x0} + a_x t$
    • Equation: $x = x_0 + v_{x0} t + \dfrac{1}{2} a_x t^2$
    • Equation: $v_x^2 = v_{x0}^2 + 2a_x \left( x - x_0 \right)$
    • Note: The equations above are written to indicate motion in the $x$-direction, but these equations can be used in any single dimension as appropriate.
  • 1.3.A.3 Near the surface of Earth, the vertical acceleration caused by the force of gravity is downward, constant, and has a measured value approximately equal to
    • Equation: $a_g = g \approx 10 \ \text{m/s}^2$.
  • 1.3.A.4 Graphs of position, velocity, and acceleration as functions of time can be used to find the relationships between those quantities.
    • 1.3.A.4.i An object's instantaneous velocity is the rate of change of the object's position, which is equal to the slope of a line tangent to a point on a graph of the object's position as a function of time.
      • Equation: $v_x = \dfrac{dx}{dt}$
    • 1.3.A.4.ii An object's instantaneous acceleration is the rate of change of the object's velocity, which is equal to the slope of a line tangent to a point on a graph of the object's velocity as a function of time.
      • Equation: $a_x = \dfrac{dv_x}{dt}$
    • 1.3.A.4.iii The displacement of an object during a time interval is equal to the area under the curve of a graph of the object's velocity as a function of time (i.e., the area bounded by the function and the horizontal axis for the appropriate interval).
      • Equation: $\Delta x = \displaystyle\int_{t_1}^{t_2} v_x(t) \ dt$
    • 1.3.A.4.iv The change in velocity of an object during a time interval is equal to the area under the curve of a graph of the acceleration of the object as a function of time.
      • Equation: $\Delta v_x = \displaystyle\int_{t_1}^{t_2} a_x(t) \ dt$

Boundary statement: AP Physics C: Mechanics and AP Physics C: Electricity and Magnetism expects that for all situations in which a numerical quantity is required for g, the value $g \approx 10 \ \text{m/s}^2$ will be used. However, students will not be penalized for correctly using the more precise commonly accepted values of $g = 9.81 \ \text{m/s}^2$ or $g = 9.8 \ \text{m/s}^2$.

Source: College Board AP Course and Exam Description

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

A displacement-time graph: the slope at any instant is the velocity A displacement-time graph: the slope at any instant is the velocity

  • 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 here.

Position, velocity, and acceleration are linked by differentiation Position, velocity, and acceleration are linked by differentiation

Explore

Explore the velocity–time graph

Change the start velocity and acceleration. The slope of a $v$–$t$ line is the acceleration $a=\tfrac{dv}{dt}$, and the area underneath is the displacement $\Delta x=\int v\,dt$ — slope and area are the graphical faces of the derivative and the integral.

Exercise sheet
1.4

Reference Frames and Relative Motion
Syllabus
Learning ObjectiveEssential Knowledge

1.4.A
Describe the reference frame of a given observer.

  • 1.4.A.1 The choice of reference frame will determine the direction and magnitude of quantities measured by an observer in that reference frame.

1.4.B
Describe the motion of objects as measured by observers in different inertial reference frames.

  • 1.4.B.1 Measurements from a given reference frame may be converted to measurements from another reference frame.
  • 1.4.B.2 The observed velocity of an object results from the combination of the object's velocity and the velocity of the observer's reference frame.
    • 1.4.B.2.i Combining the motion of an object and the motion of an observer in a given reference frame involves the addition or subtraction of vectors.
    • 1.4.B.2.ii The acceleration of any object is the same as measured from all inertial reference frames.

Boundary statement: Unless otherwise stated, the frame of reference of any problem may be assumed to be inertial.

Source: College Board AP Course and Exam Description

All motion is relative to a reference frame 参考系. Combine velocities by vector addition: $\vec{v}_{A/C}=\vec{v}_{A/B}+\vec{v}_{B/C}$. 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.

Vocabulary Train
English Chinese Pinyin
reference frame 参考系 cān kǎo xì
relative motion 相对运动 xiāng duì yùn dòng
Exercise sheet
1.5

Motion in Two or Three Dimensions
Syllabus
Learning ObjectiveEssential Knowledge

1.5.A
Describe the motion of an object moving in two or three dimensions.

  • 1.5.A.1 Motion in two or three dimensions can be analyzed using one-dimensional kinematic relationships if the motion is separated into components.
  • 1.5.A.2 Velocity and acceleration may be different in each dimension and may be nonuniform.
  • 1.5.A.3 Motion in one dimension may be changed without causing a change in a perpendicular dimension.
  • 1.5.A.4 Projectile motion is a special case of two-dimensional motion that has zero acceleration in one dimension and constant, nonzero acceleration in the second dimension.

Boundary statement: AP Physics C: Mechanics only expects students to quantitatively analyze the motion of an object in two dimensions. AP Physics C: Electricity and Magnetism expects students to also qualitatively describe the motion of a particle in three dimensions.

Source: College Board AP Course and Exam Description

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.

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.

A projectile launched at an angle: the horizontal and vertical motions are independent A projectile launched at an angle: the horizontal and vertical motions are independent

For projectile motion 抛体运动, 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. 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 $=v_{0x}\times2.0=35\ \text{m}$.

A projectile launched at an angle: velocity components, maximum height, and range A projectile launched at an angle: velocity components, maximum height, and range

Explore

Explore projectile motion

Fire the projectile, then vary the angle and speed. The horizontal motion is steady while gravity acts only downward — together they trace a parabola. Find the angle that gives the greatest range (it peaks near $45^\circ$), and try the Moon.

Vocabulary Train
English Chinese Pinyin
projectile motion 抛体运动 pāo tǐ yùn dòng
Exercise sheet
1.5

Exam tips
  • 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 and stick to it).
  • Use the dot product for work-type quantities and the cross product for torque and angular momentum.
  • Sketch the vectors — a diagram catches sign and direction errors the algebra hides.

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