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Here we will take a look at the derivation of the following kinematics equation:

The above equation solves for the displacement
of an object when it is undergoing a constant
acceleration. You need to know the ** original
velocity**, v

We will use a bit of reasoning and algebra to understand where this equation comes from.

First, let's consider the following velocity vs. time graph which shows a constant acceleration:

Since the **
slope of this graph is
the acceleration**, and since the slope of this graph
is constant, the acceleration represented here is constant.

Also, remember that the **
area under
this graph is the displacement**, or change in
position, of the object.

Now, let's consider a time period over which
this acceleration occurs. We will concern
ourselves with a *time period from t*_{1}*
to t*_{2}:

At t_{1} we will say that the
velocity is v_{1}.

And at t_{2} we will say that the
velocity is v_{2}:

Now, recall that the
area under
this graph is the displacement of the object. So the
** area** under the graph from t

Notice that this area is a **
trapezoid**, here
shaded blue:

In general, the area of a trapezoid equals one-half times the sum of the bases times the height. This trapezoid, one might say, is "on its side" since the bases are vertical and the height is horizontal.

Here is an animated diagram of a generic trapezoid of this type. This diagram demonstrates how to find the area:

Now, back to our v vs. t graph.

Notice that Base 1 is v_{1}, Base 2
is v_{2}, and that the Height is the time period from
t_{1} to t_{2}. This is shown in the
following animated diagram:

We will call the time period from t_{1}
to t_{2} to be simply t. That is:

t = t_{2} - t_{1}

So, the height of the trapezoid is t, and
the two bases are v_{1} and v_{2}. Therefore,
the area of the trapezoid is:

Area = (1/2)(v_{1} + v_{2})
t

Now, we remember that the area of a v vs. t graph is the change in position, or the displacement, and that displacement is symbolized with d.

Therefore, this:

Area = (1/2)(v_{1} + v_{2})
t

Becomes:

d = (1/2)(v_{1} + v_{2})
t

Now, as far as the acceleration over the
time period t is concerned, v_{1} is the velocity at
the start of the acceleration. That is, v_{1} is the
first or original velocity; so, the symbol v_{1} and
the symbol v_{o} mean the same thing; that is:

v_{o} = v_{1}

And v_{2} is the velocity at the
end of the acceleration. That is, v_{2} is the final
velocity, and the symbols v_{2} and v_{f} mean the same
thing:

v_{f} = v_{2}

So, this:

d = (1/2)(v_{1} + v_{2})
t

Is the same as:

d = (1/2)(v_{o} + v_{f})
t

We will now use another kinematics equation, and substitute it into our displacement equation. This other equation is:

v_{f} = v_{o} + at

See where the above equation comes from.

For the substitution we will place the
expression (v_{o} + at) into the displacement
equation where v_{f} appears. So, this:

d = (1/2)(v_{o} + v_{f})
t

Becomes this:

d = (1/2)(v_{o} + v_{o}
+ at) t

The rest is algebra:

d =
(1/2)(v_{o} + v_{o} + at) t |
Start here. |

d =
(1/2)(2v_{o} + at) t |
Since 2v_{o}
= v_{o} + v_{o}. |

d =
((1/2)2v_{o} + (1/2)at) t |
Distribute the (1/2). |

d = (v_{o}
+ (1/2)at) t |
Since (1/2)2 = 1. |

d = (v_{o}t+
(1/2)at^{2}) |
Distribute the t. |

d = v_{o}t+
(1/2)at^{2} |
Remove some parentheses, and we are done. |

We have derived our displacement equation for constant acceleration:

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