# Distance, Velocity, and Acceleration Porsche 911 image – CCA 2 Generic Do you know the relationships between distance, velocity, and acceleration?

In learning about a matter, the primary obstacle is likely to be understanding the concept of it. For instance, as a young teen, I was introduced to algebra. Algebra is the first form of mathematics that contains not only numbers, but letters as well. Numbers are constant (after all, 3 is three), but letters are used to indicate variance in value, i.e. they are variables.

## Distance

The concept of distance is so very simple, it is taken for granted. But should it be? Some young ones may not fully comprehend the concept of distance. That being said, we will assume here that the reader and his or her pupil have no difficulty understanding the concept. We will refer to distance, here, as a constant and as a variable—both. When we refer to it as a variable, we’ll use the symbol, x.

## Velocity

Velocity or speed¹ refers to the traveling of a distance over some period of time. It may be constant, or it may change. For instance, an automobile may travel a distance of 1 mile at a steady pace. This pace is said to be the velocity of the automobile. The pace is equal to the distance traveled divided by the time of travel.

When we refer to time as a variable, we will use the symbol, t. If the automobile takes 3 minutes (0.05 hours) to travel the 1 mile, then, its velocity in miles-per-hour (mph) is,

v = x / t

or

v = 1 / (0.05) = 20 mph

If that same automobile travels at that same velocity for 3 hours, the distance traveled, x, will be,

x = v • t

or

x = 20 mph • 3 hours = 60 miles

## Varying Velocity

Drivers of automobiles seldom drive them at the same speed or velocity, continuously.

For instance, Joe drives his car along a road that has a 20-mph speed limit for 2 miles. He turns right on to a road that has a 55-mph speed limit and drives the rest of the way—11 miles—to his job.

The total distance traveled, divided by the time the trip took, is, then, the average velocity.

Example: Joe travels a total of 2 miles + 11 miles = 13 miles.

The time he drives is,

(2 miles / 20 mph = 0.1 hours) + (11 miles / 55 mph = 0.2 hours)

13 miles ————- 0.3 hours

His average velocity is 13 miles / 0.3 hours ≃ 43 mph.

## Acceleration

When you drive along the road at a steady velocity, you don’t feel anything pulling or tugging your body, unless the drive accelerates (speeds up) or decelerates (slows down) by pressing down or letting up on the gas pedal (or by applying the brake).

The same effect is felt when one travels on a vigorous elevator. When it first starts up, the rider feels a force pulling him down. Gradually the force lets up. As the elevator stops, the rider feels an upward force.

Such acceleration (we will use a for acceleration) is said to be positive when it increases, and negative when it decreases (deceleration). It is a function of distance and time, both, even as velocity is. Acceleration is not quite so simple a concept as velocity is, however.

An object at rest has a velocity of zero and an acceleration of zero. So far, so good—so simple. However, an object that possesses a velocity that is not zero may still possess an acceleration of zero. Positive or negative acceleration requires a changing velocity. Let us illustrate by example.

The driver of an automobile gets into his car, turns it on, and pulls out into the road. He presses the gas pedal and reaches a velocity of 60-mph by slowly and smoothly accelerating over a period of 30 seconds. Thereafter, he maintains a steady 60-mph. What can be said about the car’s acceleration?

The equation for acceleration is

a = Δv / Δt

The triangles are actually symbols meaning change. In the case of our car, the velocity changed from 0-mph to 60-mph. So the total change, Δv = 60. Δt equals 30 seconds total in this case.² However, converting the 30 seconds to hours gives, Δt = 0.00833 hours. The acceleration is written,

a = 60 miles per hour / 0.00833 hour = 7203 mph²

Once the driver reaches 60-mph and maintains it, he is no longer accelerating. The car’s acceleration then is zero, zilch, nada.

## Using Intelligible Units

Getting back to the increasing acceleration, however, the number 7203 seems absurd and is unintelligible to most persons. This is because the driver never keeps accelerating for a full hour! We will convert it into more reasonable terms—feet-per-second-squared (ft/s²).

a = 7203 mph² • 5280 ft per mile • 1 hr per 3600 sec • 1 hr per 3600 sec

or

a = 2.93 ft/s²

This may seem ridiculously slow for a moving automobile, but consider what this number means. It means that, for during the first second, the car moves 2.93 feet. During the 2nd second, it moves 5.86 feet. During the 3rd second, it moves another 8.79 feet, and so forth.

If this still sounds low, consider a chessboard. Put a penny on square one, two pennies on square two, four pennies on square three, and so on for all 64 squares on the board. The amount of money required to accomplish the task equals \$184,467,440,737,095,516.15  See how fast the numbers add up?

## Instantaneous -vs- Average Acceleration

One last thought: Acceleration, as listed above, is actually average acceleration. In the real world, acceleration often decreases with time. In order to obtain a more accurate understanding of what goes on, the increments (changes such as the zero to 60 and the 30 seconds) should be smaller. The smaller the units, the more accurate the measure of acceleration at any particular moment.

Thus, we seek to make Δt approach zero. That is, Δt → 0. We then convert the capital Greek letter delta (Δ) into the smaller Greek delta (δ). The equation for acceleration then becomes,

a = δv / δt

So, for instance, if we calculate acceleration for every tenth of a second of the time the automobile is accelerating, we will obtain a pretty accurate mental image of what actually is going on.