In maths, algebra and graphs are often considered separately, but in Physics both algebra and graphs describe an underlying physical reality and so are just tow ways of representing, analysing and understanding a physical phenomenon.[br][br]The graphs and equations of "uniformly accelerated motion" are an excellent example of this.

To describe a property as 'uniform' is merely to say that it stays constant, so uniformly accelerated motion is simply motion with a constant rate of acceleration.

[list][*]Displacement, velocity and acceleration are vector quantities and therefore have a magnitude and a direction[/*][*]Displacement, [math]\text{\vec{s}}[/math], is the straight line distance and direction from a reference point, usually the starting point.[/*][*]Velocity, [math]\text{\vec{v}}[/math], is the rate of change of displacement with respect to time, so that the average velocity is: [br][center][math]v_{_{av}}=\frac{\Delta s}{\Delta t}[/math][/center][br][/*][*][center]Acceleration, , is the rate of change of velocity with respect to time, so that average acceleration is[/center][/*][/list][center][math]a_{av}=\frac{\Lambda v}{\Delta t}[/math][/center][br]

The simplest case of uniformly accelerated motion is when [math]a=0[/math]. In this case there is no acceleration so by definition the velocity does not change. If we plot velocity against time we will therefore get a straight horizontal line.

Prof Stick drives his kombi at a constant velocity of 2 m/s (it is a kombi after all), After 1 sec he has travelled 2 m, after two seconds he has travelled 4 m, and so on. To find the displacement, we multiple the velocity x time but notice that this gives us the same value as the area under the velocity time curve. We won't prove it here, but it turns out that this relationship is not just true for this special case of a = 0, but it holds for all cases:[size=100][b] To find the displacement from a velocity-time curve, calculate the area under the curve.[/b][/size][br][br]In cases involving straight lines we can use geometry to do this, but in more complex cases we might need to count the grid squares under the curve to estimate displacement.