The equations of motion
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 is the same for every mass.
- To handle motion like this, we use the equations of motion.
Five symbols
- Straight-line motion with constant acceleration uses five symbols.
- $u$ start velocity · $v$ final velocity · $a$ acceleration · $s$ displacement · $t$ time.
The four equations
- $v = u + at$
- $s = ut + \tfrac{1}{2}at^{2}$
- $s = \tfrac{1}{2}(u + v)t$
- $v^{2} = u^{2} + 2as$
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.
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 $v$.
Use $v = u + at = 2 + 3 \times 4 = 14\ \dfrac{\text{m}}{\text{s}}$.
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?
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.
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.
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 $g = 9.81\ \dfrac{\text{m}}{\text{s}^2}$.)
$h = \tfrac{1}{2}gt^{2} = \tfrac{1}{2} \times 9.81 \times 2.0^{2} \approx 19.6\ \text{m}$.
Thrown straight up
- The motion is symmetric: 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 $g = 9.81\ \dfrac{\text{m}}{\text{s}^2}$.)
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 remove reaction-time error.
You drop an object from several heights and plot $h$ against $t^{2}$. The gradient equals:
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 $= \dfrac{u}{g}$