The triangle symbol is the Greek letter "delta' (the uppercase version) and means 'change in'. 

Change in distance is given by final distance - starting distance; 

delta t is calculated by final time - starting time

An object will travel in a straight line at a constant speed unless some UNBALANCED force acts on it. this is Newtn's first law of motion.  Note that something ‘not moving’ has a speed of zero. 

Acceleration

Speed can change. For example, an object could speed up from 5 m s-1 to 25 m s-1 in 5 seconds.

This means it got faster by 20 m s-1 during that 5 seconds. Its acceleration is given by 

20 m s-1 ÷ 5 s =  4 m s-2

i.e. it got faster by 4 m s-1 each second: 4 metres per second per second. We can say this as “4 metres per second squared”, and we can write it as 4 m s-2.

The formula for acceleration can be written as

this can be expressed in words as "acceleration = change in speed divided by time taken"

You can use the same triangle method to rearrange this formula (there are better ways to rearrange formulae,, though)

Graphs of motion

Motion can be graphed as distance-time graphs or speed-time graphs.(acceleration-time graphs only occur at higher levels of NCEA).

• graphs you are given will only include constant accelerations over a given time interval; this means a speed -time graph will never have curves on it

• distances for NCEA level 1 are only for an object travelling in one direction, so distance-time graphs will only ever slope upwards and never downwards

• in theory, distance-time graphs cannot have sharp corners (as this would mean instant speed change); however, you may be presented with one of these if it is drawn by someone with poor physics or drawing skills.

The table below summarises what different types of line mean on a distance-time (left hand column) and speed-time (right hand column) graph.

As a general rule, speed time graphs are easier to draw than distance time graphs. The examples below will illustrate this.  Let's look at the simple speed-time graph below: For ease of description I have divided it into three sections A, B and C because in each of those sections there is a different sort of motion.

You could interpret this in words: the skater accelerated uniformly  from rest to 6 m s-1  for the first 5 seconds,  then remained at a constant 6 m s-1  for the next to seconds then decelerated (negative acceleration/slowed down)  at a constant rate to rest over the next 4 seconds.

In section A the skater is increasing speed from 0 to 6 m s-1 in a time of 5 s.

We can therefore work out an acceleration: acceleration = change in speed ÷ time taken

  = 6 m s-1 ÷ 5 s = 1. 2 m s-2

Notice a couple of things about the way I have done this.

• I have used units for all my quantities throughout the whole calculation

• i have written the units correctly (a space between the number and the units, a space between consecutive units such as metre and second,; negative index notation.

• If you were asked to describe the motion in NCEA you need to specify constant acceleration.

For section B, the gradient is zero so the acceleration is zero.

For section C, the final speed is zero and the starting speed is 6 m s-1  so the change in speed is (0-6)  = -6 m s-1 

The negative sign in this case indicates that the direction of acceleration is OPPOSITE to the direction of movement, This means that it is decelerating i,e, slowing down.

The amount of deceleration is -6 m s-1 / 4 s = -1.5 m s-2 .  Because the sign is negative, this means that the net force causing the acceleration must be backwards (opposite to the direction of movement). For example, this could be caused by the skater going from concrete to grass and hugely increasing the friction force.

Working out distance traveled on a speed-time graph

Distance on a speed time graph is given by the area under the graph. Let's go back to the graph above and look at section A. I have shaded it blue on the diagram below.

This means that by the end of Section A the skater has gone 15 m.

Section B is a rectangle, so it is easy: 6 m s-1  x  2 s = 12 m

So by the end of section B the skater has gone a total of 15 m + another 12 m

= 27 m

CAUTION: a lot of students forget to add the distances together; this is a common and avoidable exam error.

Section C is similar to section A: 0.5 x (6 m s-1  x   4 s) = 12 m

Note that the answer IS NOT NEGATIVE so the distance is STILL INCREASING.

This means that the final distance at the end of section C is the 27 m from the end of section B + the furthe 12 m while the skater was slowing down, giving a total distance for the whole journey of 39 m.

Using this information to draw a distance-time graph

This seems quite tricky but is not as hard as it looks. The main secret of doing this well is: DON'T start at the beginning (for a graph like this one)

Start with the 'steady speed' sections, because they will be straight lines on a speed time graph. So the FIRST part of the line to draw is the straight segment from 15 m to 27 m

After that, it is mainly a matter of joining the straight lines with the right sort of curve. This involves thinking: speeding up or slowing down?

If it is speeding up then the curve will be getting steeper. If it is slowing down then the curve will be getting flatter.

In the graph above, we knew the skater started at zero speed so the line starts horizontal and then gets steeper until it meets the gradient of the straight line section we have ALREADY DRAWN (if you are following my suggestion to draw this first).

To draw section C: we know the skater finished at a distance of 39 m. Place a dot at the point that represents 39 m at 11 seconds. Then, starting with the gradient you had in the straight line segment, draw a smooth flattening curve that is horizontal by the time it gets to the point you have drawn.

This takes a bit of practice; I recommend you try it a few times. A possible exam tip would be to tray drawing a few curves really lightly in pencil first, then pick your best one and go over it in pen and rub the others out. The curves in my picture are close to ideal but were done using the Bezier tool in Inkscape.

Just remember - on a distance-time graph:

If either of these things happen, have another look.

Forces

A force is a push or a pull. In most cases, an object will have more than one force acting on it, These forces add up to produce a  total force called the net force.

At Level 1 you only have to work out the net force  for forces acting in a straight line with each other. In these cases, two forces acting in the same direction add up to produce a net force that is the total of the two. 

If the two forces are in the opposite direction then the net force is in the direction of the larger force, and the size of the force is the larger force take away the smaller one. This should not be difficult, as it makes intuitive sense from many everyday experiences. The example below illustrates this:

Because they are going fast, this friction will be greater than the weight force giving a net force upwards. They will therefore accelerate upwards.

Since their motion is downwards, the upwards acceleration causes them to slow down. We can call this deceleration, but hopefully this example illustrates that so-called deceleration is really acceleration in the direction opposite to motion.

As they slow, the friction force will decrease until it once again reaches the same size as the weight force, They will once again fall at a steady speed, but this time at a much slower speed and they won't harm the environment by leaving a crater in the ground.

Although a net force of zero means an object cannot be accelerating (it must be at rest or moving in a straight line at a steady speed), applying a force to an object will cause it to change shape regardless of whether the forces are balanced or not. In many cases, such as standing on a concrete slab floor, the deformation is too small for our senses to detect (such deformation can be undesirable, which is why some very high performance bearings are made of diamond).

Before we finish with net forces, one further possible misconception needs to be debunked: many students think that an object slowing down must still have some forward force acting on it, otherwise it wouldn't be moving forward. This is untrue. 

Most of the forces you deal with in problems will be contact forces, but certain force-fields, such as gravity fields, electric fields and magnetic fields, can exert forces at a distance and through barriers.  In reality, so-called 'contact' forces (such as tension on a rope)  are just special instances of electromagnetic force, but the forces are acting at a scale our senses cannot detect.

Force-mass-acceleration relationship

You should know from previous learning that mass is a measure of the amount of substance in an object. Its units are kilograms , kg. An objects mass in kg is NOT its weight (we will get to that later).

For a given mass, the amount of acceleration it experiences is directly related to the force: twice the force, twice the acceleration, and so on. The unit of force, the newton, has therefore been defined as the amount of net force that would cause an object of 1 kg to accelerate at 1 m s-2 .

The symbol for newton is N, so we could write 1 N = 1 kg m s-2 . 

This relationship can be written in a mathematical formula as

Fnet = m a

i.e. net force (in newtons) = mass (in kilograms) times acceleration (in metres per second squared)

Like our other formulae, it can be rearranged using the triangle or other suitable methods:

• a = Fnet/m                          m = Fnet/a 

Problems involving these formulae are likely to be two-step ones that require you to calculate the acceleration then the force.

Example: A 1200 kg car travelling at 24 m s-1 brakes suddenly and comes to a halt in 3 seconds. Calculate the unbalanced force acting on the car.

Solution: step 1 is to calculate the acceleration: a = Δv/Δt  

 Δv = 0 m s-1  - 24 m s-1 = -24 m s-1 (the negative sign indicates that it is slowing down)

a = Δv/Δt= -24 m s-1/ 3 s = - 8 m s-2 

Step 2 is to use the calculated acceleration to find the force:

• F = m a  =  1200 kg x  - 8 m s-2      =  -9600 N  (again, the negative sign indicates that the force is in the opposite direction to motion)

Gravity

Gravity exerts a force of approximately 10 newtons on each kilogram of mass at the Earth's surface (the number is closer to 9.8, but 10 is the value you will be given on the NCEA exam).

This will be given to you as a value on the formula sheet as:   g = 10 N kg-1

The name for the force caused by gravity is weight. Hence, if you are given the mass of an object and are asked its weight, you need to apply the value of g and give your answer in newtons e.g.

Example 2: A sack of rice has a weight of 150 N, What is its mass?

Solution:  Fg = m g so rearranging: m = Fg / g  = 150 N / 10 N kg-1 = 15 kg

 It is common for students to be confused when they are given the weight and asked for the mass because of the everyday use of the term weight to mean mass. Try to remember the difference.

Acceleration due to gravity

The fact that gravity exerts a force on objects in direct proportion to their mass means that all objects, regardless of mass, fall with the same acceleration if there are no opposing forces such as air resistance.  Because most low-mass objects we encounter (such as a feather) are also low density and thus have a large surface area for friction, this also runs counter to our experience. If you try it with dense substances, such as lead fishing sinkers, you will find it to be true. A feather would fall just as fast - watch this video to see it happen.

As a consequence, most physicists prefer to think of gravity as an acceleration and would write

g= 10 m s-2   rather than  g = 10 N kg-1 . The two expressions are mathematically identical  but it is the newton per kilogram form you will be given on the exam paper.

Video on acceleration due to gravity            Another video with Brian Cox

Simulation - ramp and forces (requires Javascript)

The Rocket Lander

The Water Rocket       answers

In this formula, F is force in newtons and A is area in square metres.

The units of pressure are therefore newtons per square metre,  N m-2 . This is given a special name; it is called the pascal, symbol Pa.

One pascal is quite a small pressure. Many everyday pressures, such as tyre pressures, are given in kilopascals; 1 kPa = 1000 Pa.  Normal atmospheric pressure at sea level is 101 kPa.

The pressure formula can be rearranged: F = P A                   A = P/F

In many cases, you will be required to work out the area. A common mistake is to forget to convert the area to square metres or do so incorrectly.

There are three mistakes students would commonly make in the example above: they would forget to convert mass to weight; they would forget to convert centimetres to metres and they would forget to multiply by four for the four legs.

Example 2: My campervan tyres are inflated to a pressure of 310 kPa. Each of the four tyres makes a mark 15 cm x 14 cm on the ground. What is the mass of the van?

Solution: first work out the area = 0.15 m x 0.14 m x 4 = 0.084 m2

Work and energy

In science, the word work means the energy change that happens when a force moves something through a distance (the force must be in the same direction as the movement). 

The amount of work depends on the size of the force and the distance i.e.

Work = force x distance

W = F d

If force is in newtons, and distance in metres then the work is in units called joules, symbol capital J. One joule is the amount of energy you get if a force of 1 N moves a distance of 1 metre, so a joule could also be written as a newton-metre (N m).

Work and energy changes

Many of the common energy changes we deal with in science involve work. Even the energy changes in chemical reactions can involve 'work' at a fundamental level, but you would need to understand the forces acting on electrons and other exotic properties of chemical particles to think of it this way.

The main energy changes we will look at involve changes between work done and kinetic energy, or work done and gravitational potential energy, or gravitational potential energy and kinetic energy.

There is no height which is 'height zero'. So we can only specify the change in gravitational potential energy between two heights. The change in gravitational potential energy is given the symbol ΔEp and the formula becomes

ΔEp = m g Δh

Change in height is one height take away the other. The units of  ΔEp  are joules, provided the mass is in kg, the height in metres and the usual value of 10 N/kg is used for g.

The change in gravitational potential energy is the same regardless of whether an object is lifted vertically or moves horizontally as well.

In this equation

•  Ek is the kinetic energy in joules

• m is the mass in kilograms

• v is the speed in metres per second

The equation rearranges as

Example 1: A cyclist is travelling at 20 m s-1. The combined mass of the cyclist and his bike is 95 kg. Calculate the kinetic energy of the cyclist and his bike.

Solution: kinetic energy is what we are being asked, so we can use the basic equation since Ek is the subject.

Ek = ½ mv2  = ½ x 95 kg x (20 m s-1)2 = 19,000 J

Example 2: An arrow of mass 100 g has 320 J of energy. How fast is it travelling?

Solution: in this case v is the unknown so we need to use the rearrangement where v is the subject. 

Note that in both these problems, the equivalence of units is not immediately obvious; however, if you remember that a joule is a newton-metre and that a newton is a kilogram metre per second squared, you come up with the fact that a joule is a kg m2 s-2. If you were to write joules that way in your working out you would find that the units match across the equations.

Conservation of energy

The problems you are given in NCEA are quite likely to involve a change of energy from one sort into another. In many such cases, you will be expected to recognize that the total amount of energy remains the same (or nearly the same) throughout the process, and to apply that knowledge to find some unknown. Below is an example of such an energy change type problems.

Therefore the first step is to calculate the gravitational potential energy:

 ΔEp = m g Δh = 120 kg x 10 N kg-1 x 60 m  = 72,000 J

(we can easily see that the height of the ramp in this case is the Δh value)

Second step: we can now recognize that this is the kinetic energy; we have Ek and are solving for v so the equation becomes

(rounding to 2 sf as this is the most we were given in the data)

When you have finished a problem like this. always have a look and see if your answer is reasonable: 35 metres per second is a little over 120 km h-1. That seems quite fast, but not ridiculous (whereas 10 km h-1 or 500 km h-1 would definitely be suspect). In practice, some energy would be lost to friction so the skier’s actual speed would be a bit slower.

A note of interest: try repeating the problem using a different mass for the skier. Do you get the same answer? Why? (hint: play with the equations)

It is likely that the problems you will be asked to do involve two steps, either converting one form of energy to another or finding the energy a different way (such as from the power, next section). These are Excellence questions. Having the confidence to tackle them is important; if you get them right, you get E7 or E8 for the whole question (this could be a significant boost to your mark).

In this equation

• W is the amount of work done, in joules 

• t is the time taken in seconds

• P is the power in joules per second

The units for power, joules per second, are given a special name. They are called watts and the symbol for the unit is W. Please do not confuse this with the W in the equation above; the units for the W in the equation are J (yes, I know, but there are only so many letters in the alphabet).

Note: In the Electricity paper you are given this formula in a slightly different form: P = E/t

The symbol E for energy is used instead of W for work, but this is effectively the same thing. 

A point to note: students often confuse the P in this equation with the P in the pressure equation. This is almost always a sign that a student does not know what  physics ideas apply to the question they are doing. You should only need the formula sheet as a memory aid, and you should be absolutely certain for every formula what each letter stands for, what its units are, and when the formula should be used. Pressure. Power. Different things – completely different situations. Work on it until you are sure you understand the differences.

Almost all problems you have to do involving power will involve two-step working out.

This is the work done, so we need to apply the power equation: 

P = W/t = 562,500 J/5 s = 112,500 W 

(rounds to 113,000 with the significant figures given in the problem)

Again – is our answer reasonable: 113 kW is not that much more than the likely engine power for a car of this sort so seems reasonable for an acceleration of  5 m s-2  (this is half a gee).

NZQA resources here (past exam papers etc.)