Keystone High School Physics
1st Quarter - Week 2

Velocity is a vector quantity which refers to "the rate at which an object changes its position." Imagine a person moving rapidly - one step forward and one step back - always returning to the original starting position. While this might result in a frenzy of activity, it would result in a zero velocity. Because the person always returns to the original position, the motion would never result in a change in position. Since velocity is defined as the rate at which the position changes, this motion results in zero velocity. If a person in motion wishes to maximize their velocity, then that person must make every effort to maximize the amount that they are displaced from their original position. Every step must go into moving that person further from where he or she started. For certain, the person should never change directions and begin to return to the starting position.

The correct unit for displacement is meters,m and the unit for time is seconds, s.

Acceleration is a form of motion where the object�s velocity changes. It is most commonly measured in m/s/s. The formal definition for acceleration would be the rate of change of velocity with respect to time.

Stand by the last locker in a row. with your first step walk the distance of one locker. With your next step walk the distance of 2 lockers. Then, 3, 4, so on and so forth. What you are experiencing is an acceleration of one locker per step per step. This means that with each step you increase your rate by one locker per step.

The correct unit for velocity is meters per second,m/s and the unit for time is seconds, s.

Therefore the unit for acceleration is meters per second per second,m/s/s or m/s².

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 t₁to t₂:

At t₁ we will say that the velocity is v₁.

And at t₂ we will say that the velocity is v₂:

Now, recall that the area under this graph is the displacement of the object. So the area under the graph from t₁ to t₂ represents the displacement of the object over the time period from t₁ to t₂.

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.

Now, back to our v vs. t graph.

Notice that Base 1 is v₁, Base 2 is v₂, and that the Height is the time period from t₁ to t₂.

The formula for the area of the trapezoid is:

We will call the time period from t₁ to t₂ to be simply t. That is:

t = t₂ - t₁

So, the height of the trapezoid is t, and the two bases are v₁ and v₂. Therefore, the area of the trapezoid is:

Area = ½(v₁ + v₂) 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 Δx.

Therefore, this:

Area = ½(v₁ + v₂) t

Becomes:

Δx = ½(v₁ + v₂) t

Now, as far as the acceleration over the time period t is concerned, v₁ is the velocity at the start of the acceleration. That is, v₁ is the first or original velocity; so, the symbol v₁ and the symbol v_i mean the same thing; that is:

v_i = v₁

And v₂ is the velocity at the end of the acceleration. That is, v₂ is the final velocity, and the symbols v₂ and v_f mean the same thing; that is:

v_f = v₂

So, this:

Δx = ½(v₁ + v₂) t

Is the same as:

Δx = ½(v_i + v_f) t