We have talked about graphical analysis that includes finding slope of a graph… Tells us acceleration Tells us velocity t t

Another technique is to find the area under the graph. Finding the area of this rectangular section is to multiply base x height… d …But that does not tell us much for this graph t

But if you find the area under the velocity-time graph… base x height = v x t… m x s s v …you get a unit that measures displacement -very useful! t

In calculus, you call this method: finding the integral In calculus, you call this method: finding the integral. But we can use this concept without the calculus v t

Lets try another example with this same method. This graph represents an object speeding up as time ticks by. At t = 0 it has an initial velocity of v1. v v1 t

Lets try another example with this same method. …Consider the area under this graph to be broken up into a rectangle and a triangular area… Remember, that area will equal displacement v2 Area = ½ base x height v1 Area = base x height t

Adding these areas together will give us the total displacement represented by this graph… For the rectangle: A = v1 x t For the triangle: A = ½ v x t Consider: v = axt v2 v Area = ½ base x height ½ v x t + v1 v1 x t Area = base x height t

So, substituting (axt) for v in the triangle area you get: ½ (at) x t … …Or ½ a t2 For the rectangle: A = v1 x t For the triangle: A = ½ v x t Consider: v = (axt) v Area = ½ base x height ½ v x t + v1 v1 x t Area = base x height t

d = v1 t + ½ a t2 Finally, the equation we are left with is: v1 x t + For an object that has a constant acceleration v Area = ½ base x height ½ v x t + v1 v1 x t Area = base x height t

Another Proof d = v1 + v2 t v2 = v1 + at and 2 Consider two of the equations we have used for constant acceleration so far… d = v1 + v2 t and 2 v2 = v1 + at

Another Proof d = v1 + v2 t (v2 - v1) = t a and 2 Solve the second equation for time… …and substitute that new expression for time in the first equation d = v1 + v2 t and 2 (v2 - v1) = t a

Another Proof d = v1 + v2 v2 - v1 a 2a FOIL the top a 2 2a …now multiply the two fractions together

Simplify, and solve for v22 Another Proof d = v22 – v12 2a 2a d = v22 – v12 v12 + 2a d = v22 Simplify, and solve for v22