The Système Internationale d'Unités (SI units) are the base units used in measurements. The six SI units are metre, kilogram, second, ampere, kelvin and mole. These units can be combined and converted to other units, called the derived units. When very big or very small measurements are taken in SI units, they can be abbreviated using some standard prefixes. These prefixes are tera, giga, mega, kilo, cents, milli, micro and nano.

When measurements are very long, they can be rounded to an appropriate number of significant figures. The appropriate number of significant figures is usually less than or equal to the minimum number of significant figures a value in the calculations is given in.

× 10¹² = tera (T)
× 10⁹ = giga (G)
× 10⁶ = mega (M)
× 10³ = kilo (k)
× 10⁻² = centi (c)
× 10⁻³ = mili (m)
× 10⁻⁶ = micro (μ)
× 10⁻⁹ = nano (n)

A scalar quantity is one that is described only by size. A vector quantity is one that is described by both size and direction. An example of a scalar quantity is temperature, whereas an example of a vector quantity is velocity.

Distance and displacement are also examples of a scalar and a vector quantity, respectively. Distance is how far an object has travelled, whereas displacement is a measure of the total change in position from an object’s initial position.

Displacement-time graphs are graphs that show how an object’s displacement varies over time. If an object is moving, the displacement value on the graph decreases or increases. If an object is stationary, its displacement-time graph includes a horizontal line.

The area under a velocity-time graph describes the distance travelled by an object. When acceleration is uniform, the area can be calculated using the shapes under the graph. When acceleration is not uniform, the area under the graph can be calculated by counting the squares under the graph.

The area under a velocity-time graph is the distance travelled by an object. The area can be approximated by counting squares.

counting squares under graph

The distance travelled by Ophelia is given by the area under the graph.
distance = 1/2 × 6 × (60 + 150) = 630 m

630 m

12 × 10⁶ mA =
= 12 × 10⁶ × 10⁻³ A =
= 12 × 10³ A =
= 0.000012 × 10⁶ × 10³ A =
= 0.000012 × 10⁹ A =
= 0.000012 GA

0.000012 GA