Question

Tutorial: Kinematic Equations

Answer 1

Ok, I guess I'll follow the trend for giving tutorials. Since this is an introductory concept most people will be introduced to when they study physics, I guess I'll start with kinematic equations.

Answer 2

Here are the following 4 basic kinematic equations:\[x - x'= v't+1/2a*t^2\]

Answer 3

\[x -x'=1/2(v+v')t\]
\[v - v'= a*t\]
\[v^2 -v'^2 = 2as\]

Answer 4

Notation for the equations is the following:
x' = initial position
x - x' = displacement
v = velocity
v' = initial velocity
a = acceleration
s = distance covered.

Answer 5

Before going on, I'll give some detail on some of the terms presented:

Answer 6

To illustrate:

Answer 7

Now displacement can be considered as the net movement on a single axis, not the measure of net distance. Suppose, in the picture above, that a particle moved around a circle once. The net distance covered would be the circumference of the circle (2*piR), but the displacement on the x-axis and the y-axis is ZERO; the net distance traveled on both the x-axis and y-axis of the equations when the loop is complete is zero since the particle returns to its initial position.

Answer 8

Now, let's start with deriving kinematic equations.

Answer 9

Let's start in a non-calculus fashion.
To start, it is by definition that \[v = (x-x')/t\],
and \[a = (v-v')/t\]

Answer 10

With a, timing both sides by t yields \[v -v' = at\], and with x, doing the same yields \[x-x' = v*t\], or \[(x-x')/v = t\]. We'll use this shortly.

Answer 11

Now, v = (x - x')/t in this discussion is also considered as the average velocity (assuming there exist an acceleration and the acceleration is constant), which is equal to (v+v')/2, with v' being the initial velocity. In order words, the average velocity in non-calculus from can be the arithmetic mean of the initial and current velocity. From here, we equate both expressions and solve for x to get
\[x - x' = v't+1/2at^2\]

Answer 12

(Intermission: More will be added on here after I wake up in 6 hours)

Answer 13

Ok. Let's continue by comparing the two equation we've derived for x - x':\[x - x' = vt\]\[x - x' = vt +1/2at^2\]

The difference between these two equations is that the first assume that the acceleration , or change in velocity, is zero; thus in equation 1, velocity will be constant, and the displacement will proceed linearly at a constant rate, as shown on the plotss below: