Real-Life GraphsEdexcel IGCSE Maths: Revision notes

A distance-time graph plots distance travelled (y-axis) against time (x-axis).

• The gradient of the line = speed
• A horizontal line means the object is stationary (not moving)
• The steeper the line, the faster the speed
• A straight line means constant speed
• A curved line means the speed is changing (acceleration or deceleration)

Gradient formula:

$\text{speed} = \dfrac{\text{change in distance}}{\text{change in time}}$

Pick two clear points on the line, e.g. $(t_1, d_1)$ and $(t_2, d_2)$.

$\text{speed} = \dfrac{d_2 - d_1}{t_2 - t_1}$

For a straight-line section, this gives a constant speed. For a curved section, the gradient (and therefore speed) is changing at every instant — to find the speed at one exact point, you would draw a tangent to the curve at that point and find its gradient.

A negative gradient (line sloping downwards) means the object is travelling back towards its starting point.

A speed-time graph plots speed (y-axis) against time (x-axis).

• The gradient = acceleration (or deceleration if negative)
• A horizontal line means constant speed (zero acceleration)
• A straight line sloping up means constant (uniform) acceleration
• A straight line sloping down means constant deceleration

Acceleration formula:

$a = \dfrac{v - u}{t}$

where $v$ = final speed, $u$ = initial speed, $t$ = time taken. Units are usually m/s².

The area under a speed-time graph equals the distance travelled.

• For a rectangle (constant speed): area = speed × time
• For a triangle (constant acceleration from rest, or deceleration to rest): area = $\frac{1}{2}$ × base × height
• For a trapezium (e.g. speeds up, travels at constant speed, then slows down): split the shape into simpler parts (rectangles and triangles) and add the areas together, or use the trapezium area formula

Trapezium area formula:

$\text{area} = \dfrac{1}{2}(a+b)h$

where $a$ and $b$ are the parallel sides (here, the two speed values) and $h$ is the perpendicular height (here, the time interval).