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

2. 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

3. 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

4. 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

5. 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

6. 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

7. 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

8. 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

9. 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

11. 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

12. 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

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

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