Kinematics

A sprinter crosses 100 m in under 10 s, yet reaches top speed only after about 60 m. A long jumper's flight lasts less than a second, and the distance depends on a single number: the speed and angle at take-off. Kinematics turns such moments into numbers you can predict.

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 符号 (+ or −) along a chosen axis. Choosing the positive direction 正方向 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.

A vector is drawn as an arrow: its direction shows the vector's direction, and its length is proportional to the magnitude. A velocity of 10 m/s is drawn twice as long as one of 5 m/s. In equations a vector has an arrow over its symbol, $\vec v$. Along one axis we drop the arrow and let the sign carry the direction, as in $v_x$.

In this unit an object is treated as a single point, the object model: its size, shape and what is inside it are ignored. A car, a ball and a planet can all be "points" when only their overall motion matters.

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). When velocity is constant, they agree at every instant. With changing velocity, an instantaneous value can still equal the average at a particular moment. Taking the average over a shorter and shorter time interval brings it closer and closer to the instantaneous value.

An object accelerates whenever its velocity changes, in size or direction. A car going round a bend at a steady 20 m/s is accelerating, because its direction changes.

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 恒定加速度, the four kinematic equations 运动学方程 (often called SUVAT) apply:

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 末速度 and the distance it travels.

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

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):

Time to the top: $0=15-9.8\,t \Rightarrow t=1.53\ \text{s}$. By symmetry 对称性 the fall takes the same time, so the total is $3.1\ \text{s}$.

On the AP exam you may use $g\approx10\ \text{m/s}^2$; with it this ball rises 11.25 m in 1.5 s. Either value is accepted, if you use it consistently.

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

• On a position–time graph 位置–时间图, the slope 斜率 is velocity (steeper = faster; a curve = changing velocity).
• On a velocity–time graph 速度–时间图, the slope is acceleration, and the area 面积 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).

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:

Worked example (a graph that crosses the axis). A cart is pushed up a ramp and released. Its velocity is $v = 6.0 - 2.0t$ (up the ramp positive) for $5.0\ \text{s}$. Find (a) its acceleration, (b) when it turns around, (c) its displacement and (d) the distance it travels.

(a) The slope is $\dfrac{-4.0 - 6.0}{5.0} = -2.0\ \text{m/s}^2$ — the same on both sides of $t = 3\ \text{s}$. The acceleration does not change sign when the velocity does.

(b) The cart turns around when $v = 0$: $t = 3.0\ \text{s}$.

(c) Above the axis: $\tfrac12 \times 3.0 \times 6.0 = +9.0\ \text{m}$. Below: $\tfrac12 \times 2.0 \times (-4.0) = -4.0\ \text{m}$. Displacement $= 9.0 - 4.0 = +5.0\ \text{m}$ (still up the ramp from the start).

(d) Distance adds the sizes: $9.0 + 4.0 = 13\ \text{m}$.

Check (c) with $\Delta x = v_0 t + \tfrac12 a t^2 = 6.0(5.0) + \tfrac12(-2.0)(5.0)^2 = 30 - 25 = 5.0\ \text{m}$. The kinematic equation gives the displacement, never the distance, when the motion reverses.

Average velocity on a graph. The average velocity over an interval is the constant velocity that gives the same area (same displacement) in the same time. Over the 5.0 s above it is $\dfrac{+5.0}{5.0}=+1.0$ m/s. On a velocity–time graph, draw it as a horizontal line at $v=1.0$ m/s; the area under that line equals the net area under the real graph.

When the acceleration is not constant. AP Physics 1 does not ask you to calculate with changing acceleration, but you must sketch and explain it. A cart rolling over a series of bumps speeds up on each down slope and slows on each up slope: its velocity–time graph goes up and down. On a position–time graph, the slope of the tangent at a point is the instantaneous velocity; on a velocity–time graph, the slope of the tangent is the instantaneous acceleration. Where a $v$–$t$ graph gets steeper, the acceleration is growing.

Finding $a$ from data: make a straight line. A cart is released from rest at the top of a ramp, and the time $t$ to roll each distance $x$ is measured. Since $x=\tfrac12at^2$, a graph of $x$ against $t^2$ is a straight line through the origin with slope $\tfrac12a$.

1. Plot $x$ on the vertical axis and $t^2$ on the horizontal axis, with labels and units.
2. Draw a best-fit line 最佳拟合线 that passes close to all points; do not join the dots.
3. Take the slope from two points on the line that are far apart, then $a=2\times\text{slope}$. A slope of $0.60\ \text{m/s}^2$ gives $a=1.2\ \text{m/s}^2$.

Choose the axes so that the expected equation becomes $y=(\text{constant})\times x$; then the constant, the slope, contains the quantity you want.

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

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.

Observers in different inertial frames measure different velocities but the same acceleration. A ball dropped on a smoothly moving train falls straight down for a passenger and along a curve for someone on the platform, but both measure its acceleration as $g$ downward.

Worked example (one dimension). A train moves east at $20\ \text{m/s}$ relative to the ground. A passenger walks towards the back of the train at $1.5\ \text{m/s}$ relative to the train. Take east as positive.

1. $v_{P/T}=-1.5\ \text{m/s}$ and $v_{T/G}=+20\ \text{m/s}$.
2. $v_{P/G}=v_{P/T}+v_{T/G}=-1.5+20=+18.5\ \text{m/s}$: east at 18.5 m/s.
3. A car on a road beside the track moving east at $25\ \text{m/s}$ sees the passenger moving at $v_{P/C}=18.5-25=-6.5\ \text{m/s}$, that is, backwards (west) at 6.5 m/s.

AP Physics 1 keeps relative-velocity questions to one dimension. The example below, with perpendicular velocities, is extra practice with vector addition.

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:

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 $v_y=v\sin\theta$.

For projectile motion 抛体运动 (an object moving under gravity alone): the horizontal and vertical motions are independent 相互独立. 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 抛物线, and you solve it as two one-dimensional problems joined by $t$.

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 飞行时间, the maximum height 最大高度, and the horizontal range 射程 (assume it lands at launch height).

Split the launch velocity into components:

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 Horizontal motion runs at constant $v_{0x}$ for the whole flight, so the range is

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 $v_{0x}=17.3\ \text{m/s}$.

Graphs of the components. FRQs often ask you to sketch $v_x$ and $v_y$ against time for a projectile. With $g = 10\ \text{m/s}^2$ and the same launch:

Explaining in a paragraph. Some questions ask for a "paragraph-length response". Example: two spheres leave a table at the same moment, one dropped and one launched horizontally. Explain why they land together although they travel different distances.

Both spheres start with zero vertical velocity, and once they leave the table the only force on each is gravity, so both have the same downward acceleration, $g$. The vertical and horizontal motions are independent: the horizontal velocity of the launched sphere does not change how fast it falls. Starting at the same height with the same vertical velocity and acceleration, both cover the same vertical distance in the same time. The launched sphere also moves sideways at a constant speed, so it lands farther away, but at the same moment.

A good paragraph names the physics principle, applies it to this situation, and links the steps with "so" or "because", ending with the claim.

Answer each before reading the answers below. Take $g=9.8\ \text{m/s}^2$ and ignore air resistance.

1. A car moving at $12\ \text{m/s}$ brakes uniformly at $3.0\ \text{m/s}^2$ until it stops. Find the stopping time and the stopping distance.
2. A ball is dropped from rest $20\ \text{m}$ above the ground. Find the time it takes to land and its speed just before landing.
3. A ball rolls off a table $1.25\ \text{m}$ high at $8.0\ \text{m/s}$. How far from the table does it land?
4. A boat heads straight across a river at $3.0\ \text{m/s}$ relative to the water. The current flows at $4.0\ \text{m/s}$. Find the boat's speed relative to the bank.
5. A student says: "At the top of its flight, a ball thrown straight up has zero acceleration." Correct the student.

1. $t=\dfrac{12}{3.0}=4.0\ \text{s}$. Distance $=\dfrac{v^2}{2a}=\dfrac{144}{6.0}=24\ \text{m}$.
2. $t=\sqrt{2(20)/9.8}=2.0\ \text{s}$; $v=gt=9.8(2.02)=20\ \text{m/s}$.
3. Fall time $\sqrt{2(1.25)/9.8}=0.51\ \text{s}$, so it lands $8.0(0.51)=4.0\ \text{m}$ away. The horizontal speed does not change during the fall.
4. The velocities are perpendicular: $\sqrt{3.0^2+4.0^2}=5.0\ \text{m/s}$, at $\tan^{-1}(3.0/4.0)=37°$ to the bank.
5. At the top the velocity is zero for an instant, but gravity still acts, so the acceleration is still $9.8\ \text{m/s}^2$ downward. If the acceleration were zero, the ball would stay at the top.

• 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.

Common mistakes

• Saying the acceleration is zero at the top of a throw, or where a velocity–time graph crosses the axis. The velocity is zero there; the acceleration is the slope, which is not zero.
• Reading a position–time graph as if it were a velocity–time graph. Check the axis label before you use a slope or an area.
• Adding areas below the time axis as positive when the question asks for displacement. Below-axis area is negative displacement; add the sizes only for distance.
• Using $v_0 = 20\ \text{m/s}$ in the vertical equations of a projectile launched at $20\ \text{m/s}$ at an angle. Resolve first: vertical uses $v_0\sin\theta$, horizontal uses $v_0\cos\theta$.
• Using kinematic equations over a stretch where the acceleration changes (for example, braking that starts after a reaction time). Split the motion into parts with constant acceleration.
• Joining data points dot-to-dot, or taking a slope from two data points rather than from the best-fit line.
• Drawing $v_y$ for a projectile as a curve, or as a V shape that bounces at the top. It is one straight line with slope $-g$ that passes through zero.
• Giving $g$ a sign that does not match your positive direction. With up positive, $a = -9.8\ \text{m/s}^2$ on the way up and on the way down.