Difficulty understanding formula and area under v-t graph

[aboojoo]

1. The problem


For sake of format I attached the a screenshot of the course material I'm having difficulty wrapping my walnut around. Which is how:

Total Displacement = Area of Triangle + Area of Rectangle
or
Δvector d = A_triangle + A_Rectangle
or
Δvector d = 1/2 (V₂-V₁)Δt +V₁*Δt

Translates in to:

Δvector d= 1/2 (V₁+V₂)Δt


Could someone explain it to me? I feel as if I'm blatantly missing an obvious answer as to why this is but since there is no answer in the text, I'm reaching out.

help.png contains screenshot
q.doc contains my question in better formatting

 

[PhanthomJay]

Is it the algebra you are having trouble with? Or the concept?
1/2k( a -b) + kb =
1/2ka - 1/2kb + kb =
1/2ka. + 1/2kb =
1/2k(a + b)

Conceptually, for constant acceleration , displacement is average velocity times change in time, where average velocity is the sum of the initial and final velocities all divided by 2.

 

[NascentOxygen]

Hi aboojoo! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

If velocity were constant, then the graph would be a horizontal line, and you'd be happy to multiply that velociy value by Δt to find the displacement during that interval of time, Δt. You can see that by performing that multiplication you are actually finding the area of a rectangle, the area under that graph. The same property holds even where the velocity is not constant; you could think of dividing it up into lots of thin vertical rectangles if you wish, where velocity is almost constant during each narrow time interval. However you look at it, you are still determinng the area under the curve during that whole time interval. Once you appreciate this, you can use whatever method is easiest to determine the area under the graph, e.g., a large rectangle with a triangular piece atop it, as illustrated in this problem.

 

[aboojoo]

I totally understand now , thanks a lot guys you've really helped, both conceptually and algebraic.