Kinematics

A speedometer shows speed: the distance travelled per unit time.

These five quantities come up in almost every kinematics 运动学 question. Learn the exact words — the examiner gives marks for precise wording.

• distance 距离 — the total length of the path travelled. A scalar 标量.
• displacement 位移 — the straight-line distance from the start to the end, with a direction. A vector 矢量.
• speed 速率 — the rate of change of distance with time. A scalar.
• velocity 速度 — the rate of change of displacement with time. A vector.
• acceleration 加速度 — the rate of change of velocity with time. A vector.

The unit of speed and velocity is $\text{m s}^{-1}$; the unit of acceleration is $\text{m s}^{-2}$.

A common mistake: deceleration 减速度 just means acceleration in the opposite direction to the velocity. It is not a separate quantity.

A high-speed train: its motion can be shown on a distance-time graph.

Many marks come from reading or drawing motion graphs. Two graphs matter.

Displacement–time graph

The gradient 斜率 (steepness) of a displacement–time graph at a point gives the velocity at that moment.

• flat line → the object is at rest.
• straight sloping line → constant velocity (gradient = velocity).
• curved line → changing velocity. Draw a tangent 切线 at the point and find its gradient.

Displacement–time graph of a car on a test track

Velocity–time graph

The gradient of a velocity–time graph gives the acceleration at that moment.

The area between the line and the time axis gives the displacement in that time.

• flat line → constant velocity (zero acceleration).
• straight sloping line → uniform acceleration 匀加速 (constant acceleration).
• curved line → changing acceleration.
• area above the time axis is positive displacement; area below is negative (the object moved backwards).

Velocity–time graph — gradient gives acceleration, area gives displacement

Acceleration–time graph derived from the same motion

To find the displacement, split the area into triangles and rectangles, or count grid squares. Area of a triangle is $\tfrac{1}{2} \times \text{base} \times \text{height}$; area of a rectangle is $\text{base} \times \text{height}$.

For motion in a straight line with uniform acceleration, we use five symbols: starting velocity $u$, final velocity $v$, acceleration $a$, displacement $s$, and time $t$. Four equations link them:

Each equation uses four of the five symbols. To pick the right one: write down what you know and what you want, then choose the equation with exactly those four.

Worked example. A car accelerates uniformly from $8\ \text{m s}^{-1}$ to $20\ \text{m s}^{-1}$ over a distance of $56\ \text{m}$. Find its acceleration.

We know $u$, $v$ and $s$ and want $a$, so use $v^{2} = u^{2} + 2as$:

Where the SUVAT equations come from

You should be able to get these from the definitions of velocity and acceleration:

• $v = u + at$ comes from $a = (v - u)/t$, the gradient of the line.
• $s = \tfrac{1}{2}(u + v) t$ is the area under the line — a trapezium 梯形 with parallel sides $u$ and $v$ and width $t$.
• $s = ut + \tfrac{1}{2}at^{2}$ comes from putting $v = u + at$ into the area.
• $v^{2} = u^{2} + 2as$ comes from removing $t$ from the first two.

Displacement–time graph for uniform acceleration — the slope at any point equals the instantaneous velocity

If a question asks "which equation can be found using only the gradient of a velocity–time graph?", the answer is $v = u + at$ (the gradient is the acceleration).

Choosing a positive direction

Pick a positive direction at the start and keep it. Anything pointing the other way gets a minus sign. For a ball thrown straight up, if "up" is positive: $u$ is positive, $a = -g$ (gravity 重力 pulls down), and at the highest point the displacement is positive but the velocity is zero.

When air resistance 空气阻力 can be ignored, an object in free fall 自由落体 has a constant acceleration $g \approx 9.81\ \text{m s}^{-2}$ downwards. This is the same for every mass.

For a ball dropped from rest and falling a distance $h$:

For a ball thrown straight up with speed $u$:

• greatest height: put $v = 0$ in $v^{2} = u^{2} - 2gh$, giving $h = u^{2}/(2g)$.
• time to reach the top: put $v = 0$ in $v = u - gt$, giving $t = u/g$.
• total time to fall back to the start height: $2u/g$ (the motion is symmetric 对称).

Worked example. A ball is thrown straight up at $20\ \text{m s}^{-1}$. Find the greatest height it reaches (take $g = 9.81\ \text{m s}^{-2}$).

At the highest point $v = 0$, so from $h = u^{2}/(2g)$:

Experiment to find $g$

A common method: drop an object from rest, then measure the distance $h$ it falls and the time $t$ it takes. Then

Repeat for several heights and plot $h$ against $t^{2}$. The gradient of the best straight line is $g/2$, so $g$ is twice the gradient. Repeating reduces random error 随机误差. An electronic timer — using light gates 光电门, or a switch the ball hits — removes reaction-time 反应时间 error.

Experimental set-up for measuring the acceleration due to free fall

Water jets from a sprinkler trace parabola paths — a real example of projectile motion

When an object moves at constant velocity in one direction (say horizontal 水平) and speeds up in a direction at right angles to it (say vertical 竖直, under gravity), the two motions do not affect each other. Treat each direction on its own, with its own SUVAT equation.

Horizontal throw

An object thrown horizontally with speed $u_{\text{H}}$ from height $h$, with air resistance ignored:

• horizontal: constant velocity $u_{\text{H}}$. After time $t$, the horizontal distance is $x = u_{\text{H}} t$.
• vertical: starts from rest and speeds up downwards at $g$. After time $t$, it has fallen $y = \tfrac{1}{2} g t^{2}$ and has vertical velocity $v_{\text{V}} = g t$.

The time to reach the ground depends only on the height $h$, not on $u_{\text{H}}$. Solve $h = \tfrac{1}{2} g t^{2}$ for $t$; then the horizontal range 射程 is $u_{\text{H}} t$.

Worked example. A ball is thrown horizontally at $15\ \text{m s}^{-1}$ from the top of a cliff $20\ \text{m}$ high. Find the time it takes to land and how far from the base it lands (take $g = 9.81\ \text{m s}^{-2}$).

Vertical motion gives the time: from $h = \tfrac{1}{2}g t^{2}$,

Horizontal motion then gives the range: $x = u_{\text{H}} t = 15 \times 2.0 \approx 30\ \text{m}.$

The horizontal-velocity graph is a flat line at $u_{\text{H}}$. The vertical-velocity graph is a straight line from the origin with gradient $g$.

Projectile at an angle

A projectile 抛体 thrown at speed $u$ at angle $\theta$ above the horizontal:

Projectile launched at angle $\theta$ — horizontal and vertical motions are independent

• horizontal component 分量 of the starting velocity: $u_{\text{H}} = u \cos\theta$ (stays constant during the flight).
• vertical component of the starting velocity: $u_{\text{V}} = u \sin\theta$ (gets smaller, becomes zero at the top, then grows downwards).

At the highest point, $v_{\text{V}} = 0$, but $v_{\text{H}}$ is still $u\cos\theta$. The time to the top is $t_{\text{up}} = u\sin\theta / g$; the total flight time (back to the start height) is $2t_{\text{up}}$.

Range $R$ of a projectile launched from and landing on level ground

Bouncing ball

When a ball bounces, its velocity–time graph is a set of straight sloping lines (constant $g$) with a sudden jump at each bounce (the velocity flips direction, and gets smaller if some energy 能量 is lost). Add up the times and the distances across the bounces.

When two objects move along the same line in different ways, write a displacement equation for each. Use the same start time and the same positive direction. Then set the two displacements equal (or set their difference to a given gap).

For a goods train at constant velocity $u_{\text{G}}$ and an express train starting from rest with acceleration $a$, both passing the same point at $t = 0$:

They are level again when $s_{\text{G}} = s_{\text{E}}$, giving $t = 2 u_{\text{G}} / a$.

1. Draw a diagram and mark the positive direction.
2. List the SUVAT symbols with their known and unknown values, including signs.
3. Choose the SUVAT equation with exactly the four symbols you have, plus the one you want.
4. For projectile motion, split into horizontal and vertical SUVAT problems, linked only by the time $t$.
5. Always check the units of your answer, and that its size is sensible.

• Use the SUVAT equations only for constant acceleration; list $s, u, v, a, t$ and pick the equation missing your unknown.
• Choose one direction as positive and keep signs consistent (usually $g = -9.81\ \text{m s}^{-2}$).
• On a velocity-time graph, gradient $=$ acceleration and area $=$ displacement.
• For projectiles, treat horizontal (constant velocity) and vertical ($a = g$) motion separately, linked by the same time.