**What is acceleration?**Most people more or less know this. But, let's start with an example!

To begin, imagine sitting in a car at a red light. When the light turns green, the driver presses the gas pedal. WHICH is CALLED the accelerator because… Pressing the accelerator, the flow of fuel to the engine increases causing the RPMs of the engine to increase.

The passengers of the car feel themselves pressed back into the seat as the car moves forward. The speedometer begins to show a change in speed!

5 MPH

10 MPH

20 MPH

40 MPH

That image is more than is needed to understand the concept of acceleration!

*Acceleration is the rate that velocity changes over some period of time.*

The effect of acceleration is to change velocity. This can be explored mathematically quite easily.

**EXAMPLE:**

Suppose a car is rolling along a highway at 55 MPH and the speed limit changes to 65. How much did it accelerate (speed up)?

Inspection of the numbers answers that question. It speed up by 10 MPH.

Inspection of the numbers answers that question. It speed up by 10 MPH.

So… using

*vf*for final velocity,*vi*for initial velocity, and*∆d*for how much velocity changed:
vf = 65 mph

vi = 55 mph

∆v = 10 mph

The equation is equally intuitive:

vf = vi + ∆v

Suppose we wanted to find vf?

Find vf where

vi = 55 mph

∆d = 10 mph

vf = vi + ∆v

vf = 55 mph + 10 mph

vf = 65 mph

(Of course, you could solve for any of the numbers!)

**There is a strong similarity between changing velocity and changing position.**

When an object changes position, the rate of that change is velocity. When an object changes velocity, the rate of that change is acceleration. It is a whole lot like what has been learned in previous articles about velocity and position.

*To quickly review that…*

To find how far an object moves—that is to say, to find the magnitude of an object's displacement—the rate of position change is multiplied by the amount of time that it changes. Or, as an equation,

df = vt

wheredis change in position (displacement),vis rate of change in position, (average velocity), andtis elapsed time.

In some cases, displacement is not measured from the point at which motion begins, but is measured from some other fixed point. In those cases, the total displacement becomes the sum of the initial displacement (initial position) and the distance moved. In such a case,

df = di + vt

wheredfis the final displacement from the fixed point,diis the initial displacement from the fixed point,vis rate of change in position, (average velocity), andtis elapsed time. (And where elapsed time the difference between final time and initial time.)

**Finding Final Velocity**

*What is true for displacement closely parallels what is true for acceleration.*
The change in velocity can be found by multiplying the rate that velocity changes (acceleration rate) by the amount of time that it changes (elapsed time). As an equation,

v = at

where

vis the change in velocity (sometimes written as ∆v),

ais the rate of acceleration, and

tis the elapsed time.

In many cases, an object is already in motion (has an initial velocity) when the acceleration begins. Logic and reason lead to the conclusion that the final velocity of an object is the initial velocity plus or minus the change in velocity. Since change in velocity is found (see above) easily, the equation for final velocity is

vf = vi + at

where v

*f*is final velocity,*vi*is initial velocity,*a*is the rate of acceleration, and*t*is the elapsed time. (And where elapsed time the difference between final time and initial time.)**In many cases, however,**

*Vi*is**0 (zero)**.

**Finding Average Velocity**

To extend and begin to combine the above principles of motion, it is necessary to first consider how to find average velocity when given the rate of acceleration and time. Fortunately, this is relatively logical and easy to reason out. And the math on it is equally easy!

To apply logic, if something accelerates at a constant rate, it starts our at one velocity and moves faster and faster until it reaches a final velocity. Suppose it starts out a 0 m/s and ends up at 4 m/s. Through the course of its acceleration, it passes through all of the velocities between 0 and 4 m/s.

0 sec, v = 02 sec, v = 14 sec, v = 26 sec, v = 38 sec, v = 4

The data above shows and elapsed time of 8 seconds, and in that time, the velocity of the object changed from 0 to 4 m/s. It might be tempting to just guess that, since the final velocity was 4 m/s, over the 8 seconds, the average velocity was 2 m/s. If that guess was made, the answer would be correct.

The list below gives

*Vf, Vi*, and Vave:Vf= 0, Vi=6, Vave= 3

Vf= 10, Vi=16,, Vave= 13

Vf= 0, Vi=10, Vave= 5

Vf= 100, Vi=160, Vave= 130

Vf= 4, Vi=8, Vave= 6

Simply allowing logic and reason to stand alone would violate the need of science to describe everything with equations! So, there is an equation:

vave = (vf+ vi) /2

**So what if v**

*i***is zero**? The math gets even easier!

vave = (vf+ 0) /2

vave = vf/2

**NOTES AND HINTS:**

See also THIS.

**The Equations:**

*df = (di + ∆d)*Use this if:

- You need to find the final distance, total distance, or final position AND You know BOTH di and how much d changes (e.g. it moves 10 meters)

*vf = vi + ∆v*Use this if:

- You need to find the final velocity
ANDYou knowBOTHvi and how much v changes (e.g. speed increases by 10 m/s)

**v(ave) = (vi + vf)/2**Use this if:

- You need to find the average velocity
- You know BOTH vi and vf (you might have to calculate vf!)

**This will answer almost all “how fast” questions.**

*vf = vi + (a)(t)*Use this if:

- 2 velocities are given
- You don’t have distance

**This will answer almost all “how far” questions.**

*df = di + (vi)(t) + (½)(a)(t*^{2})Use this if:

- Only 1 velocity is given
- You have a distance given (df is rarely 0, but sometimes it is the unknown)

**The hints and helps:**

It is

**vi**if…

“…traveling at a rate of…” “…moving at…”"…has a velocity of…"

If it is “at rest” then vi = 0 and di is probably 0

The words "begin" and "starts" generally go with the initial values.

If it is “moving at a constant” rate or if it “has a constant velocity” then a = 0

It is possible that something not given (but which is not the thing to be found) should have a value of 0 (zero).

**Moving on to Displacement**

The ability to find an average velocity allows more work with acceleration to be done. Since, if the average velocity is known, the displacement can be found. The work can be done in three steps.

- First, find the final velocity.
- Next, find the average velocity.
- Last, use the average velocity and the elapsed time to find the displacement.

**However, doing the work in three steps is not necessary, because math.**

**The best, most direct, preferred process for the calculation of final distance/displacement will be revealed HERE.**

*Yet, just for the sake of having completed the thoughts above, an example is provided.*

**EXAMPLE**:

A car with an initial velocity of 20 meters per second (m/s) accelerates at a uniform rate of 3 (m/s)/s for 10 seconds. How far does it travel while it is accelerating?Step 1.)Find the final velocity.vf = vi + atvf = 20 + 3 • 10vf = 20 + 30vf = 50 mphStep 2.)Find the average velocity.v(ave) = (vf + vi)/2v(ave = (50 + 20)/2v(ave) = 70/2v(ave) = 35 mphStep 3.)Find the displacement. NOTE: the elapsed time is the same for displacement and acceleration.df = di + vtdf = 0 + 35 • 10df = 350 meters (350 m)

By combining an understanding of average velocity with the ability to find displacement and to find final velocity, it becomes possible to calculate the displacement that occurs during a period of acceleration. This 3-step method can be used when given acceleration and time and asked to find distance.

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