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Kinematics

AP Physics 1 · Topic 1

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Handout Vocabulary
1.1

Scalars and Vectors in One Dimension
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.3.i Vectors are notated with an arrow above the symbol for that quantity.
      • Equation: $\vec{v} = \vec{v}_0 + \vec{a}t$
    • 1.1.A.3.ii Vector notation is not required for vector components along an axis. In one dimension, the sign of the component completely describes the direction of that component.
      • Derived equation: $v_x = v_{x0} + a_x t$

1.1.B
Describe a vector sum in one dimension.

  • 1.1.B.1 When determining a vector sum 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 how objects move, without asking why. First, two kinds of quantity:

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

The difference matters. Distance is the total path length travelled – a scalar that only grows. Displacement is the straight-line change in position, with a direction. Walk $3\ \text{m}$ east then $1\ \text{m}$ back west: the distance is $4\ \text{m}$, but the displacement is only $2\ \text{m}$ east.

In one dimension, direction is just a sign (+ or −) along a chosen axis. Choosing the positive direction first is essential – every vector's sign depends on it. A velocity of $-5\ \text{m/s}$ does not mean "slow"; it means $5\ \text{m/s}$ in the negative direction.

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
distance 距离 jù lí
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: $\vec{v}_{avg} = \dfrac{\Delta \vec{x}}{\Delta t}$
  • 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: $\vec{a}_{avg} = \dfrac{\Delta \vec{v}}{\Delta t}$

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

  • 1.2.B.4 An object is accelerating if 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.

Source: College Board AP Course and Exam Description

Projectile motion is two independent motions

Three linked vectors describe motion along a line:

  • Displacement 位移 $\Delta x$ is the change in position – a vector from start to end (not the total path length, which is distance).
  • Velocity 速度 is the rate of change of position, $v=\dfrac{\Delta x}{\Delta t}$. Its sign gives direction; its magnitude is speed 速率.
  • Acceleration 加速度 is the rate of change of velocity, $a=\dfrac{\Delta v}{\Delta t}$.

Be careful to separate average velocity 平均速度 (total displacement over total time) from instantaneous velocity 瞬时速度 (the velocity at one instant, the slope of the position–time graph at that point). They are equal only when the velocity is constant.

An object speeds up when $v$ and $a$ have the same sign, and slows down (deceleration 减速) when they have opposite signs. Note that a negative acceleration does not always mean slowing down – a ball falling faster and faster has negative velocity and negative acceleration.

For constant acceleration, the four kinematic equations (often called SUVAT) apply:

$$v=v_0+at,\qquad \Delta x=v_0 t+\tfrac{1}{2}at^2,\qquad v^2=v_0^2+2a\,\Delta x,\qquad \Delta x=\tfrac{1}{2}(v_0+v)\,t.$$
Pick the equation that contains the three quantities you know plus the one you want, so only one unknown is left. They apply only while $a$ is constant.

Worked example. A car starts from rest and accelerates uniformly at $2.0\ \text{m/s}^2$ for $6.0\ \text{s}$. Find its final velocity and the distance it travels.

List what you know: $v_0=0$, $a=2.0\ \text{m/s}^2$, $t=6.0\ \text{s}$.

$$v=v_0+at=0+2.0\times 6.0=12\ \text{m/s},$$
$$\Delta x=v_0 t+\tfrac12 at^2=0+\tfrac12\times 2.0\times 6.0^2=36\ \text{m}.$$

Worked example (free fall). A ball is thrown straight up at $15\ \text{m/s}$. Taking $g=9.8\ \text{m/s}^2$ and up as positive, how high does it rise, and how long is it in the air before returning to the thrower's hand?

At the highest point the velocity is momentarily zero, and $a=-g=-9.8\ \text{m/s}^2$ throughout (this is free fall 自由落体, ignoring air resistance 空气阻力):

$$v^2=v_0^2+2a\,\Delta x \;\Rightarrow\; 0=15^2+2(-9.8)\Delta x \;\Rightarrow\; \Delta x=\frac{225}{19.6}=11.5\ \text{m}.$$
Time to the top: $0=15-9.8\,t \Rightarrow t=1.53\ \text{s}$. By symmetry the fall takes the same time, so the total is $3.1\ \text{s}$.

Vocabulary Train
English Chinese Pinyin
Displacement 位移 wèi yí
Velocity 速度 sù dù
speed 速率 sù lǜ
Acceleration 加速度 jiā sù dù
average velocity 平均速度 píng jūn sù dù
instantaneous velocity 瞬时速度 shùn shí sù dù
deceleration 减速 jiǎn sù
free fall 自由落体 zì yóu luò tǐ
air resistance 空气阻力 kōng qì zǔ lì
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: $v_x^2 = v_{x0}^2 + 2a_x(x - x_0)$
    • 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 $a_g = g \approx 10 \ m/s^2$.
  • 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.
    • 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.
    • 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).
    • 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.

Boundary statement: AP Physics 1 does not expect students to quantitatively analyze nonuniform acceleration. However, students will be expected to be able to qualitatively analyze, sketch appropriate graphs of, and discuss situations in which acceleration is nonuniform.

Boundary statement: For all situations in which a numerical quantity is required for $g$, the value $g \approx 10 \ m/s^2$ 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

The same motion appears as a description, a graph, a table, or an equation, and you should move between them:

Reading a distance-time graph: flat means at rest, a straight slope means constant speed Reading a distance-time graph: flat means at rest, a straight slope means constant speed

  • On a position–time graph, the slope is velocity (steeper = faster; a curve = changing velocity).
  • On a velocity–time graph, the slope is acceleration, and the area under the line is displacement.

Reading slopes and areas off graphs is a core exam skill. To get displacement from a velocity–time graph, split the area into triangles and rectangles and add them up; area below the time axis counts as negative displacement (motion the other way).

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

Worked example. A cyclist speeds up uniformly from rest to $8.0\ \text{m/s}$ in $4.0\ \text{s}$, then holds $8.0\ \text{m/s}$ for $6.0\ \text{s}$. Find the total distance from the velocity–time graph.

The area is a triangle followed by a rectangle:

$$\Delta x=\underbrace{\tfrac12\times 4.0\times 8.0}_{\text{triangle}}+\underbrace{6.0\times 8.0}_{\text{rectangle}}=16+48=64\ \text{m}.$$

Explore

Explore the velocity–time graph

Change the start velocity $u$ and the acceleration $a$. The gradient (slope) of the line is the acceleration; the area between the line and the time axis is the displacement.

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.

Boundary statement: Adding or subtracting vectors to find relative velocities is restricted to motion along one dimension for AP Physics 1.

Source: College Board AP Course and Exam Description

All motion is measured against a reference frame 参考系. Velocities measured in different frames differ, and you combine them by vector addition. The velocity of A relative to C is

$$\vec{v}_{A/C}=\vec{v}_{A/B}+\vec{v}_{B/C}.$$
A person walking on a moving train has one velocity relative to the train and another relative to the ground – this is relative motion 相对运动. A useful shortcut: the velocity of A relative to B is $\vec{v}_{A/B}=\vec{v}_A-\vec{v}_B$ (subtract B's velocity).

A frame that is not accelerating is an inertial reference frame 惯性参考系: one in which a free object (no net force) obeys Newton's first law, staying at rest or moving at constant velocity. A frame that accelerates – a braking car – is non-inertial, where objects seem to speed up with no force acting on them.

Worked example. A boat points straight across a river and moves at $3.0\ \text{m/s}$ relative to the water. The current flows at $4.0\ \text{m/s}$ along the river. Find the boat's speed and direction relative to the bank.

The two velocities are perpendicular, so add them as a right triangle:

$$v=\sqrt{3.0^2+4.0^2}=5.0\ \text{m/s},\qquad \theta=\tan^{-1}\!\frac{4.0}{3.0}=53^{\circ}\ \text{downstream from straight across.}$$

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

Vectors and Motion in Two Dimensions
Syllabus
Learning ObjectiveEssential Knowledge

1.5.A
Describe the perpendicular components of a vector.

  • 1.5.A.1 Vectors can be mathematically modeled as the resultant of two perpendicular components.
  • 1.5.A.2 Vectors can be resolved into components using a chosen coordinate system.
  • 1.5.A.3 Vectors can be resolved into perpendicular components using trigonometric functions and relationships.
    • Equation: $\sin \theta = \dfrac{a}{c}$
    • Equation: $\cos \theta = \dfrac{b}{c}$
    • Equation: $\tan \theta = \dfrac{a}{b}$
    • Equation: $a^2 + b^2 = c^2$

1.5.B
Describe the motion of an object moving in two dimensions.

  • 1.5.B.1 Motion in two dimensions can be analyzed using one-dimensional kinematic relationships if the motion is separated into components.
  • 1.5.B.2 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.

Source: College Board AP Course and Exam Description

In two dimensions, resolve each vector into components 分量 along perpendicular axes ($x$ and $y$), handle each axis separately, then recombine. A velocity $v$ at angle $\theta$ to the horizontal has components $v_x=v\cos\theta$ and $v_y=v\sin\theta$.

A velocity vector resolved into its horizontal and vertical components A velocity vector resolved into its horizontal and vertical components

For projectile motion 抛体运动 (an object moving under gravity alone): the horizontal and vertical motions are independent. Horizontally, velocity is constant ($a_x=0$); vertically, acceleration is $-g$ (down). The two motions share only the time. So a projectile's path (its trajectory 轨迹) is a parabola, and you solve it as two one-dimensional problems joined by $t$.

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

A dropped ball and a horizontally launched ball fall together – the vertical motions are identical A dropped ball and a horizontally launched ball fall together – the vertical motions are identical

Worked example. A ball is kicked at $20\ \text{m/s}$, $30^{\circ}$ above the horizontal. Taking $g=9.8\ \text{m/s}^2$, find the time of flight, the maximum height, and the horizontal range 射程 (assume it lands at launch height).

Split the launch velocity into components:

$$v_{0x}=20\cos 30^{\circ}=17.3\ \text{m/s},\qquad v_{0y}=20\sin 30^{\circ}=10\ \text{m/s}.$$
Vertical motion sets the time. At the top $v_y=0$, so $0=10-9.8\,t \Rightarrow t_{\text{up}}=1.02\ \text{s}$, and the total flight is $2t_{\text{up}}=2.0\ \text{s}$. The maximum height is
$$\Delta y=\frac{v_{0y}^2}{2g}=\frac{10^2}{19.6}=5.1\ \text{m}.$$
Horizontal motion runs at constant $v_{0x}$ for the whole flight, so the range is
$$R=v_{0x}\times t_{\text{flight}}=17.3\times 2.0=35\ \text{m}.$$

A common trap: at the top of the flight the vertical velocity is zero, but the ball is not at rest – its horizontal velocity $v_{0x}$ never changes. The speed at the top equals $v_{0x}=17.3\ \text{m/s}$.

Explore

Explore projectile motion

Fire the ball, then change the angle and speed. The horizontal motion stays steady while gravity pulls it down — together they trace a parabola. Find the launch angle that gives the longest range, and try the Moon.

Explore

Explore vectors and their components

Drag the vectors to change their $x$- and $y$-components. See how a single vector is built from independent horizontal and vertical parts, and how two vectors add tip-to-tail into a resultant.

Vocabulary Train
English Chinese Pinyin
components 分量 fèn liàng
projectile motion 抛体运动 pāo tǐ yùn dòng
trajectory 轨迹 guǐ jì
range 射程 shè chéng
Exercise sheet
1.5

Exam tips
  • Choose the right kinematic equation by listing the three quantities you know plus the one you want, so only one unknown remains; the SUVAT equations apply only while acceleration is constant.
  • Fix a positive direction first — every displacement, velocity, and acceleration then carries a sign; a negative velocity means "moving the other way", not "slow".
  • Treat a projectile as two independent 1-D problems sharing only the time $t$: constant velocity horizontally, $a=-g$ vertically. At the top $v_y=0$ but $v_x$ is unchanged.
  • On a velocity–time graph the gradient is the acceleration and the area is the displacement (area below the axis is negative).
  • Distinguish distance (scalar, total path) from displacement (vector, start-to-end), and speed from velocity.

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