Why does the kinematic equation use t² for the triangular area under acceleration?

[Matt1234]

Homework Statement

See the attachment please.

The Attempt at a Solution

I know the v1 x delta (t) is for the rectangle,
I don't see how the second part works? I don't understand why the t^2 and how a replaces the area formula i have written.


Thanks for your help, its my first day of class today, and its been a while since i was last in a physics / math class.

EDIT: I added a formula sheet we were given, sorry i just don't see the relationship right now..

 

[CompuChip]

You can replace $\Delta v = v_2 - v_1$ by the acceleration a. In the picture, the acceleration is the slope of the line, algebraically it is given by
$$a = \frac{\Delta v}{\Delta t}$$.

So just write down the area of the triangle and do the substitution.

 

[wavingerwin]

from the second equation on the formula sheet,
a=$$\Delta$$v/$$\Delta$$t
a=(v_f-v_i)/t

from the graph, d=v_it+$$\frac{1}{2}$$(v_f-v_i)*t
therefore, d=v_it+$$\frac{1}{2}$$at²
(since v_f-v_i=at)