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 shows speed: the distance travelled per unit time
The equations of motion
v = u + at
The suvat results come straight from the v–t graph: gradient = a, area = displacement.
The four equations
- $v = u + at$
- $s = ut + \tfrac{1}{2}at^{2}$
- $s = \tfrac{1}{2}(u + v)t$
- $v^{2} = u^{2} + 2as$
Acceleration–time graph derived from the same motion
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.
Displacement–time graph for uniform acceleration — the slope at any point equals the instantaneous velocity
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.
Match each equation to the quantity it does not contain.
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Knowing which quantity each equation leaves out is what makes the choice instant.
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 m/s. What is the greatest height it reaches, in metres? (g = 9.81 m/s^2)
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At the top v = 0, so v^2 = u^2 + 2as gives 0 = 400 - 2(9.81)s and s = 20.4 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 $t = \dfrac{u}{g} = \dfrac{20}{9.81} \approx 2.0\ \text{s}$.
Measuring $g$
- Drop an object a height $h$ and time the fall: $g = \dfrac{2h}{t^{2}}$.
- 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
- 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 $= \dfrac{u}{g}$