Uniform motion

1. In uniform motion an object moves in a straight line at a constant speed, so it covers the same distance in every second.
2. Distance, speed and time for uniform motion are linked by $distance\; travelled=speed\times time$, an equation you need to recall.
3. In symbols this is $s=v\times t$, where $s$ is the distance in $m$, $v$ is the speed in $m/s$ and $t$ is the time in $s$.
4. Rearranging gives $v=\frac{s}{t}$ to find a speed and $t=\frac{s}{v}$ to find a time.
5. To rearrange, divide both sides of $s=v\times t$ by the quantity you want to remove, so dividing both sides by $v$ leaves $t=\frac{s}{v}$.
6. Every value is converted into SI units before it is substituted, so a time in minutes is changed into seconds and a distance in kilometres into metres.
7. For uniform motion the distance-time graph is a straight line, and its gradient is the constant speed $v$.

Uniform acceleration

1. In uniform acceleration the velocity changes by the same amount every second, so the velocity-time graph of the motion is a straight line.
2. Acceleration is linked to the change in velocity and the time by $acceleration=\frac{change\; in\; velocity}{time}$, an equation you need to recall.
3. The change in velocity is the final velocity $v$ minus the initial velocity $u$, so in symbols the equation is $a=\frac{v-u}{t}$.
4. Acceleration is measured in ${m/s}^2$, because it is a change in velocity in $m/s$ divided by a time in $s$.
5. An acceleration of $3{m/s}^2$ means that the velocity increases by $3m/s$ in every second.
6. When an object slows down, $v$ is smaller than $u$, so the acceleration comes out negative and is called a deceleration.
7. The phrase "starts from rest" tells you that $u=0m/s$, and the phrase "comes to rest" tells you that $v=0m/s$.
8. Rearranging gives $v=u+a\times t$ to find the final velocity and $t=\frac{v-u}{a}$ to find the time.
9. When the time is not known, a second equation links the two velocities, the acceleration and the distance, $v^2-u^2=2\times a\times s$.
10. You do not need to memorise $v^2-u^2=2as$, but you must be able to select it and apply it.
11. Rearranging it gives $a=\frac{v^2-u^2}{2s}$, $s=\frac{v^2-u^2}{2a}$ and $v=\sqrt{u^2+2as}$.
12. To choose an equation, list the quantities you know and the one you need, then pick the equation that contains exactly those quantities.
    1. If $u$, $v$ and $t$ are involved, use $a=\frac{v-u}{t}$.
    2. If $u$, $v$ and $s$ are involved and no time is given, use $v^2-u^2=2as$.

Measuring acceleration

1. An acceleration is found from two velocities and the time between them, so an experiment has to measure the speed at two points as well as the time taken to travel between them.
2. An interrupt card of known length is fixed upright on a trolley, so that it blocks the beam of a light gate as it passes through.
3. The speed at a light gate is the length of the card divided by the time the beam is blocked, $speed=\frac{length\; of\; card}{time\; beam\; is\; blocked}$.
4. With two light gates, the speed at each gate and the time between the gates give the acceleration from $a=\frac{v-u}{t}$.
5. A double interrupt card with two tabs gives two speeds from a single light gate, and the data logger records the time between the two tabs.
6. With one light gate and a trolley released from rest, $u=0$, so $v^2-u^2=2as$ gives the acceleration from the speed at the gate and the distance from the release point.
7. A ticker timer is an older method that prints dots on a paper tape at fixed time intervals, so dots that become further apart show a trolley speeding up.
8. Light gates are preferred because the blocking times are often less than a tenth of a second, which is far too short to measure by hand.