The equations of motion
Introduced| English | Meaning |
|---|---|
| free fall/friː fɔːl/ | Motion under gravity alone, with other forces neglected. |
| equations of motion/ɪˈkweɪʒnz ɒv ˈməʊʃn/ | equations of motion |
| constant acceleration/ˈkɒnstənt əkˌseləˈreɪʃn/ | Motion in which velocity changes by equal amounts over equal time intervals. |
| symmetric/sɪˈmetrɪk/ | Having corresponding parts or relationships that remain equivalent under a specified transformation or interchange. |
| light gate/laɪt ɡeɪt/ | A timing sensor that detects an object interrupting a beam of light. |
A hammer and a feather
- On the Moon, an astronaut dropped a hammer and a feather together — they landed at the same time.
- With no air, free fall 自由落体 is the same for every mass.
- To handle motion like this, we use the equations of motion 运动方程.
Five symbols
- Straight-line motion with constant acceleration 匀加速 uses five symbols.
- $u$ start velocity · $v$ final velocity · $a$ acceleration · $s$ displacement · $t$ time.
A speedometer gives the size of the velocity only; in SUVAT, $u$ and $v$ also carry a sign for direction
The equations of motion
v = u + at
The suvat results come straight from the v–t graph: gradient = a, area = displacement.
The four equations
- $v = u + at$
- $s = ut + \tfrac{1}{2}at^{2}$
- $s = \tfrac{1}{2}(u + v)t$
- $v^{2} = u^{2} + 2as$
- They hold only while the acceleration is constant. If $a$ changes, split the motion into stages or use a graph.
When may the equations of motion be used? Select all that apply.
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Air resistance makes the acceleration change, so only graph methods work there. A curved velocity-time graph is the visible warning.
Choosing the right one
- Each equation uses four of the five symbols.
- Write down what you know and what you want, then pick the equation with exactly those.
Each SUVAT equation leaves out one symbol. Match each to the one it does not contain.
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Spotting the missing symbol is how you choose: pick the equation that has your four known/wanted symbols and leaves out the one you neither know nor want.
A car starts at $u = 2\ \dfrac{\text{m}}{\text{s}}$ and accelerates at $a = 3\ \dfrac{\text{m}}{\text{s}^2}$ for $t = 4\ \text{s}$. Find $v$.
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Use $v = u + at = 2 + 3 \times 4 = 14\ \dfrac{\text{m}}{\text{s}}$.
Put the method for choosing a suvat equation in order.
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Picking the equation that is missing the quantity you neither have nor want saves solving two equations together.
Worked example: a car brakes to a stop
- A car travelling at $20\ \dfrac{\text{m}}{\text{s}}$ brakes steadily and stops in $50\ \text{m}$. Known: $u = 20\ \dfrac{\text{m}}{\text{s}}$, $v = 0$, $s = 50\ \text{m}$. Wanted: $a$. Time is neither known nor wanted, so use $v^{2} = u^{2} + 2as$.
- $a = \dfrac{v^{2} - u^{2}}{2s} = \dfrac{0 - 20^{2}}{2 \times 50} = -4.0\ \dfrac{\text{m}}{\text{s}^2}$. The minus sign means the acceleration points against the motion. Then $t = \dfrac{v - u}{a} = \dfrac{0 - 20}{-4.0} = 5.0\ \text{s}$.
A car travelling at $25\ \dfrac{\text{m}}{\text{s}}$ brakes steadily and stops in $40\ \text{m}$. What is the size of its deceleration?
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Time is not given or wanted, so use $v^{2} = u^{2} + 2as$. $a = \dfrac{0 - 25^{2}}{2 \times 40} = -7.8\ \dfrac{\text{m}}{\text{s}^2}$; the size is $7.8\ \dfrac{\text{m}}{\text{s}^2}$.
Where they come from
- $v = u + at$ is the gradient of a velocity–time line ($a = \tfrac{v-u}{t}$).
- $s = \tfrac{1}{2}(u+v)t$ is the area under it (a trapezium).
Which equation is just the gradient of a velocity–time graph?
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The gradient is $a = \dfrac{v - u}{t}$, which rearranges to $v = u + at$.
Pick a positive direction first and keep it. For a ball thrown up with "up" positive, $a = -g$ — gravity points the other way.
A ball is thrown straight up at $20\ \dfrac{\text{m}}{\text{s}}$. What is the greatest height it reaches? (Use $g = 9.81\ \dfrac{\text{m}}{\text{s}^2}$.)
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At the top $v = 0$, so $v^{2} = u^{2} + 2as$ gives $0 = 20^{2} + 2(-9.81)s$ and $s = 20.4\ \text{m}$. With up positive, $a$ is negative for the whole flight.
Free fall
- Ignoring air, a falling object accelerates at $g \approx 9.81\ \dfrac{\text{m}}{\text{s}^2}$ downward — the same for every mass.
- Dropped from rest: $h = \tfrac{1}{2}gt^{2}$, $\;v = gt$, $\;v^{2} = 2gh$.
With no air resistance, a heavy ball falls faster than a light one.
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No — free fall has the same acceleration $g$ for every mass. Air resistance is what makes a feather seem to fall slower.
A stone is dropped from rest and falls for $2.0\ \text{s}$. How far does it fall? (Use $g = 9.81\ \dfrac{\text{m}}{\text{s}^2}$.)
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$h = \tfrac{1}{2}gt^{2} = \tfrac{1}{2} \times 9.81 \times 2.0^{2} \approx 19.6\ \text{m}$.
Thrown straight up
- The motion is symmetric 对称: it rises, stops, and falls back the same way.
- Greatest height $h = \dfrac{u^{2}}{2g}$; time to the top $t = \dfrac{u}{g}$; total time up and down $= \dfrac{2u}{g}$.
A ball is thrown straight up at $20\ \dfrac{\text{m}}{\text{s}}$. How long until it reaches the top? (Use $g = 9.81\ \dfrac{\text{m}}{\text{s}^2}$.)
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At the top $v = 0$, so $t = \dfrac{u}{g} = \dfrac{20}{9.81} \approx 2.0\ \text{s}$.
Worked example: thrown up from a cliff
- A ball is thrown up at $15\ \dfrac{\text{m}}{\text{s}}$ from a cliff top $20\ \text{m}$ above the sea. Take up as positive: $u = +15\ \dfrac{\text{m}}{\text{s}}$, $a = -9.81\ \dfrac{\text{m}}{\text{s}^2}$, and the sea is below the start, so $s = -20\ \text{m}$. One equation covers the whole flight; do not split it at the top.
- $s = ut + \tfrac{1}{2}at^{2}$ gives $-20 = 15t - 4.905t^{2}$, so $4.905t^{2} - 15t - 20 = 0$ and $t = 4.1\ \text{s}$ (reject the negative root). Then $v^{2} = u^{2} + 2as = 15^{2} + 2(-9.81)(-20)$, so $v = -25\ \dfrac{\text{m}}{\text{s}}$: $25\ \dfrac{\text{m}}{\text{s}}$ downward.
A stone is thrown straight up at $10\ \dfrac{\text{m}}{\text{s}}$ from the top of a cliff $15\ \text{m}$ high. How long is it until the stone hits the ground at the foot of the cliff? (Use $g = 9.81\ \dfrac{\text{m}}{\text{s}^2}$.)
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With up positive: $u = +10\ \dfrac{\text{m}}{\text{s}}$, $a = -9.81\ \dfrac{\text{m}}{\text{s}^2}$, $s = -15\ \text{m}$. $s = ut + \tfrac{1}{2}at^{2}$ gives $4.905t^{2} - 10t - 15 = 0$, so $t = 3.0\ \text{s}$.
Two stages, two sets of SUVAT
- When the acceleration changes, each stage gets its own $u, v, a, s, t$. The final velocity of one stage is the start velocity of the next.
- A car starts from rest and accelerates at $3.0\ \dfrac{\text{m}}{\text{s}^2}$ for $4.0\ \text{s}$: $v = 0 + 3.0 \times 4.0 = 12\ \dfrac{\text{m}}{\text{s}}$ and $s = \tfrac{1}{2} \times 3.0 \times 4.0^{2} = 24\ \text{m}$. It then brakes at $5.0\ \dfrac{\text{m}}{\text{s}^2}$: $0 = 12^{2} + 2(-5.0)s$ gives $s = 14.4\ \text{m}$, so it travels $38\ \text{m}$ in total.
The acceleration jumps from $+3$ to $-5$ at time $t$: two constant-acceleration stages, each solved with its own SUVAT
A train starts from rest and accelerates at $0.50\ \dfrac{\text{m}}{\text{s}^2}$ for $20\ \text{s}$. It then moves at a constant speed for $30\ \text{s}$. How far does it travel in total?
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Stage 1: $v = 0 + 0.50 \times 20 = 10\ \dfrac{\text{m}}{\text{s}}$ and $s = \tfrac{1}{2} \times 0.50 \times 20^{2} = 100\ \text{m}$. Stage 2 starts at $10\ \dfrac{\text{m}}{\text{s}}$ with $a = 0$: $s = 10 \times 30 = 300\ \text{m}$. Total $400\ \text{m}$.
Measuring $g$
- Drop an object a height $h$ and time the fall: $g = \dfrac{2h}{t^{2}}$. A ball falling $1.20\ \text{m}$ in $0.495\ \text{s}$ gives $g = \dfrac{2 \times 1.20}{0.495^{2}} = 9.8\ \dfrac{\text{m}}{\text{s}^2}$.
- Plot $h$ against $t^{2}$ — the gradient is $\dfrac{g}{2}$. Light gates 光电门 remove reaction-time error.
You drop an object from several heights and plot $h$ against $t^{2}$. The gradient equals:
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From $h = \tfrac{1}{2}gt^{2}$, plotting $h$ against $t^{2}$ gives a straight line of gradient $\tfrac{1}{2}g$ — so $g$ is twice the gradient.
You've got it
- four SUVAT equations link $u, v, a, s, t$ — pick the one with the four symbols you have
- they need constant acceleration; a change of acceleration starts a new stage
- choose a positive direction: a displacement below the start, or $g$ when up is positive, is negative
- free fall: constant $g \approx 9.81\ \dfrac{\text{m}}{\text{s}^2}$ down, the same for every mass
- a vertical throw is symmetric: time up $= \dfrac{u}{g}$