Acceleration, Velocity, Distance, and Time

The study of motion with relationship to time considers primarily how the position or velocity of an object changes with respect to time. In exploring these changes, definitions and formulas emerge:

displacement: the magnitude of the change in position of an object.

velocity: the rate at which the position of an object changes with respect to time.

acceleration: the rate at which the velocity of an object changes with respect to time.

df = di + vt

where df is final (or total) displacement, di is initial displacement, v is average velocity, and t is elapsed time.

vf = vi + at

where vf is final velocity, vi is initial velocity, a is acceleration, and t is elapsed time.

v(ave) = (vf+vi)/2

where v(ave) is average velocity, vf is final velocity, and vi is initial velocity.

Using the formulas above, it is possible to directly solve problems that ask about any combination of displacements, velocity and time or that ask about velocities, acceleration and time. However, to solve problems that ask about displacements, acceleration, and time, three steps were needed, if only the above formulas were used.

To solve the problems that that ask about displacements, acceleration, and time directly, a fourth formula is needed. This formula can be derived from the other three, but to get started, here it is:

df = di + vit + 1/2at²          (t² means t•t or "t squared")

This equation is much more fun with di and vi are zero! But, to break it down as is first

df is the final, total displacement.

di is the initial displacement. How far from whatever point of reference is the object when the thing starts accelerating?

vit accounts for the motion of the object based on its starting velocity. It keeps covering distance at the initial rate.

The last term tells how much MORE distance is covered based on the acceleration. Since any di and vi are covered in the first term, the last term can be analyzed for the zero case, but first two examples.

EXAMPLE 1 (the zero case)

How far will a rocket travel while it accelerates from rest at a rate of 4 (m/s)/s for 5 seconds?

di = 0

vi = 0

vf = not given, not asked for

df = what you are looking for

a = 4

t = 5

df = di + vit + 1/2at²

df = 0 + 0 + 1/2 • 4 • 5•5

PEMDOS

df = 1/2 • 4 • 25

df = 2 • 25

df = 50

EXAMPLE 2

How far will a rocket travel while it accelerates from 20 m/s at a rate of 6 (m/s)/s for 3 seconds?

di = 0

vi = 20

vf = not given, not asked for

df = what you are looking for

a = 6

t = 3

df = di + vit + 1/2at²

df = 0 + 20•3 + 1/2•6 • 3•3

PEMDOS

df = 60 + 1/2 • 6 • 9

df = 60 + 3 • 9

df = 60 + 27

df = 87

CONCLUSION:

The above examples can be checked using the three-step method. The equation used above can be used to solve for any of the variables, but does require dealing with squares and square roots. However, if that level of math does not present a problem, 

df = di + vit + 1/2at²

becomes the only equation needed for all cases of motion related to displacement. Combined with

vf = vi + at

all of the motion problems can be solved.

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Example 3 (Find a when given distance and time.)

Working through the three-step method to find acceleration for the case of di and vi = 0 results in a quick and easy calculation method that requires

Step 1: Divide df by t

Step 2: Double it

Step 3: Divide by t again.

Using the new equation, this emerges like this:

df = 1/2 a t²

df = 1/2 a • t • t

2 • df = a • t • t

2df/t = a • t

(2df/t)/t = a

(Same as the 3 step method! Wooo!)

To more correctly write the above equation results in:

2•(df/t²) = a

And that is called physics!

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WHO LOVES SOME MATH?

Okay, don't answer that. 

It is not hard to come up with the df = di + vit + 1/2at² formula. Starting with the basic, simple case formulas, combining them quickly results in

df = 1/2at²

How is that done? Easily! Begin with

df = vt 

where v is the average velocity.

Remember that 

vf = at 

where vf is the final velocity. In the case of vi = 0, the formula for average velocity easily reduces:

v(ave) = (vf + vi)/2

v(ave) = (vf + 0)/2

v(ave) = (vf)/2

Now, back to vf = at! Plug it in!

So…

v(ave) = (vf)/2 becomes

v(ave) = (at)/2

Now, back to df = vt! Plug in (at)/2 for v:

df = (at)/2 • t

Clean up time:

df =  a • 1/2 • t • t

df = 1/2 • a • t • t

df = 1/2at²

Bam! That just happened!

Displacement and Acceleration

If the relationship between displacement, acceleration, and elapsed time is extended to the most general case, the following equations need to be combined.

1.)

df = di + vt

where df is the final displacement from the fixed point, di is the initial displacement from the fixed point, v is rate of change in position, (average velocity), and t is elapsed time.

2.)

Vf = Vi + at

where Vf is final velocity, Vi is initial velocity, a is the rate of acceleration, and t is the elapsed time.

3.)

Vave = (Vf  + Vi) / 2

where Vf is final velocity, Vi is initial velocity, and Vave is the average velocity.

To begin, start with the general displacement equation:

df = di + vt

Understanding that v is average velocity, the equation becomes:

df = di + ((Vf  + Vi) / 2) • t

And since Vf can be found in relationship to acceleration, the following emerges:

df = di + (((Vi + at) + Vi) / 2) • t

Solving the problem in steps generally is more easily understood. To find displacement when acceleration is present, do the following.

Step 1.) Find the final velocity: Vf = Vi + at

Step 2.) Find the average velocity: Vave = (Vf  + Vi) / 2

Step 3.) Find the displacement: df = di + vt

EXAMPLE

A model car is 3 meters from the starting line on a model car race track and it is moving at 2 m/s. It accelerates at a rate of 5 (m/s)/s for 4 seconds. How far does it end up from the starting line?

Step 0.) Collect the data!

di = 3 m

vi = 2 m/s

a = 5 (m/s)/s

t = 4

Step 1.) Find final velocity.

Vf = vi + at

vf = 2 + 5•4

vf = 2 + 20

vf = 22

Step 2.) Find average velocity.

v(ave) = (vf + vi)/2

v(ave) = (22 + 2)/2

v(ave) = 24/2

v(ave) = 12 m/s

Step 3.) Find the total displacement.

df = di + vt

(remember v is average velocity and t is the same elapsed time as for the acceleration)

df = 3 + 12•4

df = 3 + 48

df = 51 meters 

So, the model car ends up 51 meters from the starting line.

Concepts of Acceleration

Introduction to Chemistry and Physics Index

Physics Index

Concepts of Acceleration

Before starting an exploration of acceleration, it is useful to revisit how we think about velocity. Previously, velocity was described like this:

average velocity: the value represented by the displacement of an object in relationship to the elapsed time. Remember that displacement is the measure of the length of a line lying directly between the initial position and a final position of an object that moved.

What is key in this is that velocity is the rate at which the position of an object changes. The equation that relates displacement to velocity and time is:

d = vt

where d is displacement, v is average velocity, and t is elapsed time.

To complicate things (because that's what we do) displacement can be re-imaged as a change of position—the difference between initial position and final position—and elapsed time can be re-imaged as the difference between initial time and final time.

Okay, so if complicated formulas intimidate you, it's okay to close your eyes for a few seconds!

The re-imaged equation becomes, where:

di = initial position (or beginning displacement from some reference point)

df = final position (or final displacement from some reference point)

ti = initial time (clock reading at the beginning, when the object is at di)

tf = final time (clock reading at the end, when the object is at df)

v = average velocity 

(df  - di) = v ( tf - ti)

Okay, you can open your eyes now!

The "ah ha moment" above is that velocity relates to how the position of the object changes. Using words instead of math symbols, the idea is a lot more logical.

Velocity is the rate that position changes over some period of time.

This could be stated another way with displacement being the subject. Think about it this way:

The change in position of an object is determined by its rate of motion (velocity) and how long it moves.

Further, if it is some distance from a reference point to begin with (di), it will end up that distance plus how far it moved. That is evident in this equation:

df = di + vt

(back to that, yeah…)

Bam!

Say that about 10 times!

On To Acceleration!

Now, with the understanding of velocity refined, understanding acceleration becomes much simpler. There is a perfect parallel between acceleration concepts and velocity concepts!

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, the RPMs 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 parallels between acceleration and velocity carry through to the math equations as well. Just to prove the point, the previous discussion of velocity is going to be copied below this sentence and all the relevant word changes will be made in red!

What is key in this is that acceleration is the rate at which the velocity of an object changes. The equation that relates velocity to acceleration and time is:

vf = vi + at

where v is change in velocity, a is rate of acceleration, and t is elapsed time.

To complicate things (because that's what we do) change in velocity can be re-imaged as the difference between initial velocity and final velocity, and elapsed time can be re-imaged as the difference between initial time and final time.

Okay, so if complicated formulas intimidate you, it's okay to close your eyes for a few seconds!

The re-imaged equation becomes, where:

vi = initial velocity (or beginning velocity from some reference point)

vf = final velocity (or final velocity from some reference point)

ti = initial time (clock reading at the beginning, when the object is at di)

tf = final time (clock reading at the end, when the object is at df)

a = acceleration rate

(vf  - vi) = a ( tf - ti)

Okay, you can open your eyes now!

The "ah ha moment" above is that acceleration relates to how the velocity of the object changes. Using words instead of math symbols, the idea is a lot more logical.

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

This could be stated another way with change in velocity being the subject. Think about it this way:

The change in velocity of an object is determined by its rate of acceleration and how long it moves. Or, to save pixels:

v = at

(back to that, yeah…)

So you can see that acceleration and velocity are similar, but that velocity relates to the change of position and acceleration relates to the change of velocity.


Another parallel concept between acceleration and velocity is when the object has an initial velocity and then undergoes acceleration. This is just like, with displacement, when an object is some distance from a starting point and begins to move. For velocity, the equation is:

df = di + vt

where df is the final distance from a reference point, di is the initial distance from a reference point, v is average velocity and t is elapsed time.

The relationship between final velocity and acceleration is found using a parallel equation:

vf = vi + at

where vf is the final velocity, vi is the initial velocity, a is acceleration rate and t is elapsed time.

In many cases, the "final velocity" concept is easier to visualize than is "final distance." Look at the two examples that follow.

Final Distance

A car starts out 4 miles from the state line and travels at an average velocity of 20 miles per hour for 2 hours. How far from the state line does it end up?

For the above, the equation is (how far does it end up?):

df = di + vt

df = 4 mi + 20mi/hr • 2 hr

df = 4 mi + 40mi

df = 44 mi

While this is not terribly complicated, processing the idea of "how far from the state line" is a bit unfamiliar. The idea that the car moves 20 MPH for 2 hours is fairly simple—40 miles is traveled. So, the car moves 40 miles from the initial position, which is 4 miles from the state line, so… Well, in the end, it seems a little less obvious.

HOWEVER, with acceleration, the idea of final velocity is more readily accessible. The basis of the idea is that something is moving at a given velocity ( ) and either speeds up or slows down. Thus, it has a new, different velocity. Hence:

vf = vi + at

Final Velocity

A car starts out traveling at 40 mi/hr. It speeds up at a constant rate of 10 mph/sec for 2 second. How fast does it end up traveling?

For the above, the equation is (How fast does it end up traveling?):

vf = vi + at

vf = 40 mi/hr + 10 (mi/hr)/sec • 2 sec

vf = 40 mi/hr + 20 mi/hr

So the change in speed is + 20 mi/hr. Thus, the final velocity is…

vf = 60 mi/hr

In many cases, the initial velocity will be zero. This will be true whenever the situation includes an object "at rest" or "not moving." Here are some phrases to look for that allow you to assume the initial velocity is zero:

An object is at rest…Two stationary zebras…The airplane begins to move… 

Generally, unless there is some definite indication of an initial velocity, you can assume vi = 0 and if so, the equation reduces nicely to:

vf = at

However, it is good practice to start with the complete equation each time, fill in the give values, and then solve. One more step is not going to kill you!