How can we derive the 'suvat' equations of motion v=u+at and s=(u+v)t/2
- v=u+at Using a velocity time graph with constant acceleration, (graph provided as visual aid in tutorial) with initial velocity u and final velocity v, we can see that the graph is a straight line. It therefore takes the form y=mx+c (the straight line equation), where velocity v is on the y axis, time t is on the x axis and initial velocity u is the y intercept c. This therefore shows that acceleration a is given by the gradient of the graph, and this can be proved using the definition of acceleration: 'rate of change of velocity' or 'change in velocity divided by time taken'. (Rate = divided by time taken). Here change in velocity is given by v-u so a=(v-u)/t which can be rearranged (rearrangement shown in detail in tutorial) to give v=u+at. 2) s=(u+v)t/2 This can be shown simply by knowing that displacement = velocity x time. But we need to take the average velocity given by (u+v)/2 and then multiply this by time t. This can also be shown using the velocity time graph from before. Since displacement is given from change in velocity x change in time we can multiply the axes of the graph to get displacement, which can therefore be found from the area under the graph. To calculate this area we simply use the formula for the area of a trapezium. ("Half the sum of the parallel sides times the distance between them, that's the way you calculate the area of a trapezium!"), which again gives s=(u+v)t/2 from the graph. (Graph again shown in more detail in the tutorial). We have successfully derived the first two (and arguably most important) suvat equations of motion! The three further suvat equations can then be simply derived from these two :) Thanks for joining me!
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