Ugh okay guys, kinematics lesson.

Deriving the kinematics equations from the definition of uniform acceleration

__doesn't require calculus at all.__ **I recommend every MCAT student to derive them themselves so they can memorize them forever.** ** Of course do this before the MCAT.** It takes maybe 20 minutes.

vf = final velocity, vo = initial velocity

Let's start with the definition of acceleration: vf = vo + at (this one you have to learn by rote memorization). Since this is a linear equation, we can graph this. You should know that the slope of this graph is the acceleration and the area under the curve is displacement (d).

(1) vf = vo + at

Now let's figure out how we can use geometry to figure out the area to get displacement.

The area of the bottom rectangle is

height*

base or

vo*

t.

The area of each triangle is 0.5*

height*

base or 0.5*

at*

t or 0.5at².

The area of the overall rectangle (from the top red line to the time axis) is

height*

base or

vf*

t.

The area under the curve is therefore:

the bottom rectangle +

the bottom triangle or

(2) d =

vo*t +

0.5at²
the overall rectangle -

the top triangle or

(3) d =

vf*t -

0.5at²
Now what if we chop the top half of the triangle and put on the other side? We would get a rectangle whose area would be equal to the original area (see picture below).

The area of this rectangle is

height*

base or

(4) d =

0.5*(vo + vf)*

t
The last equation is tricky. We have to plug in the first equation into the fourth equation. From equation (1), we rearrange to

**t = (vf - vo)/a** and plug t into equation (4).

d = 0.5*(vo + vf)*

**(vf - vo)/a** or

(5) d = 0.5*(vf² - vo²)/a

Let's look at the MEANING behind each equation.

(1) vf = vo + at

This is the definition of acceleration. It means, that acceleration is equal to the change in velocity over time.

(2) d =

vo*t + 0.5at²

What if we take out "0.5at²"? What does "

vo*t" physically mean? That is just distance = rate * time for an object that is traveling at constant initial velocity. But since the object is accelerating, it will go an EXTRA distance. So we have to add the "0.5at²" in.

(3) d =

vf*t - 0.5at²

"

vf*t" would be the displacement if the object was traveling at the final velocity the entire time. But it didn't, it traveled at a slower velocity and accelerated. So we have to subtract some displacement because it wasn't going that fast the entire time. Thus, we have to subtract "0.5at²" out.

(4) d =

0.5*(vo + vf)*t

What is "

0.5(vo + vf)"? That's the

average velocity the object is traveling. So this equation really says "displacement =

average velocity * time" which is an equation you already know.

(5) d = 0.5*(vf² - vo²)/a

This is just the restatement of equation (4). Multiplying the average velocity by time gives us the displacement. In this case, time = (vf - vo)/a which simplifies into this.

By understanding where the equations come from and what the meaning is behind each equation, you not only have an easier time remembering each equation, but you have the deeper understanding required to do well on the MCAT.

(Created with MS Paint)