Area and distance travelled

1. The area between a velocity-time graph and the time axis represents the distance travelled by the object during that time.
2. This works because distance equals velocity multiplied by time, and on the graph the velocity is the height of the area while the time is its width.
3. The units confirm the idea, because a velocity in $m/s$ multiplied by a time in $s$ gives a distance in $m$.
4. For a constant velocity the line is horizontal, so the area underneath it is a rectangle with a height equal to the velocity and a width equal to the time.
   1. A car moving at a steady $15m/s$ for $6s$ covers $15\times 6=90m$, which is the area of the rectangle under its line.
5. For a uniform acceleration from rest the line slopes up from the origin, so the area underneath it is a triangle.
   1. The area of a triangle is $\frac{1}{2}\times base\times height$, where the base is the time and the height is the final velocity.
6. For a uniform acceleration between two velocities the area is a trapezium, which can be split into a rectangle under the starting velocity and a triangle on top of it.
   1. The trapezium can also be found in one step as $\frac{1}{2}\times (initial\; velocity+final\; velocity)\times time$.
7. A journey with several stages is split into sections, the area of each section is found, and the areas are added to give the total distance.
8. The gradient and the area give two different quantities from the same graph, because the gradient gives the acceleration while the area gives the distance.

Curved lines and counting squares

1. When the line is curved, the area under it cannot be split exactly into simple shapes, so it is estimated by counting the squares of the grid.
2. First work out the distance that one grid square represents by multiplying the velocity of its height by the time of its width.
3. Next count every whole square under the line, count each square that is half or more under the line as one square, and ignore each square that is less than half under it.
4. Multiplying the number of squares by the distance per square gives an estimate of the total distance.
5. A finer grid gives a better estimate, because less of the area is wrongly gained or lost along the edge of the curve.
6. Another method replaces the curve with a few straight lines joining points on it, which turns the area into triangles and trapezia that are quick to calculate.

Areas below the time axis

1. Velocity is a vector, so a section of a velocity-time graph below the time axis shows the object moving in the opposite direction.
2. An area above the axis represents a displacement in the positive direction, and an area below the axis represents a displacement in the negative direction.
3. The total distance travelled is found by adding the sizes of all the areas, treating every area as positive.
4. The overall displacement is found by giving areas below the axis a negative sign before adding, so these areas cancel part of the areas above the axis.
5. For example, a ball rolled up a slope that rolls back down past its starting point could have an area of $6m$ above the axis and an area of $9m$ below it.
6. The ball then travels a total distance of $6+9=15m$ but ends with a displacement of $6-9=-3m$, which places it $3m$ below its starting point on the slope.