Basic Units of Motion
Distance and time are realities for all mobile creatures.
Most creatures deal with these realities intuitively.
Humans learned to consciously plan for them.
This "Prequel to the Thought Experiment" briefly reviews distance, time, velocity (distance per time), and acceleration (velocity per time).
DISTANCE
Distance: Basic Thots. Most people are familiar with English units of distance. Examples include:
 Inch  probable measure of average man's forefinger tip.
 Foot  probable measure of average man's foot.
 Yard  length of a typical's man's stride.
Metric. About the time of the French Revolution, two French astronomers endeavoured to create a scientific standard of distance. (See Ken Adler's book, The Measure of All Things, a highly enjoyable read.) After years of pain staking and highly precise measurements, they determined a unit of measure now called the meter. Thus, a meter closely approximates one millionth of the distance between North Pole and Equator. In grade school, we learn that a centimeter (cm) is 1 hundredth of a meter, and one inch is equal to 2.54 cm.
Recall from middle school  1 in = 2.54 cm 
Multiply by ten  10 in = 25.4 cm 
Double and add  +2 in = +5.08 cm 
Total  12 in = 30.48 cm 
Multiply by 5000  5000 ft = 1525 m 
× 200  200 ft = 61 m 
× 80  80 ft = 24.4 m 
Total of above  5,280' = 1,610.4 m = 1.61km 
TIME
Day is defined by the duration of one rotation of the Earth .
Man has divided this period such that:
1 day = 24 hours = 86,400 seconds 
1 min = 60 sec 

1 hour = 60 min
1 hr = 60 * 60 sec = 3,600 sec
1 day = 24 hrs = 86,400 sec
Thus, time is slightly more consistent then distance which must convert between different units of same systems as well as between different systems of measurement. However, all such conversions can be done with simple arithmetic operations.VELOCITY
Velocity equals distance over time. (v = d/t) Via a simple procedure, one can readily convert velocities between different dimensions (example: from Int'l System's km/hr to British System's ft/sec and vice versa).
Sidebar: Is speed different from velocity?? 
Academically, velocity is a vector quantity with both direction and amount, while speed is a scalar quantity which is only an amount. However, this article follows common convention which disregards this distinction and assumes same meaning. 
Recall that 3 feet is a little less then 1 meter. (3 ft = .915 m = 1 English yard). Recall that a yard was chosen as the length of average man's stride; thus, a brisk walking pace could cover about 60 yards per min = 3,600 yards per hour. Since 3,600 yards = 3 × 3,600 ft = 10,800'; then, one can readily estimate a brisk walk can easily cover 2 statue miles per hour.
Thus, a walking pace of 1 m/s will exceed 2 miles/hr. Let's convert 1 meter per second (Int'l System) to equivalent miles per hour (British System).
Convert to meters per hr.  1 m/sec × (3,600s/hr) = 3,600 m/hr 
Convert to km per hr.  3,600 m/hr = 3.6 km/hr 
Recall 5 mi = 8 km  1 m/s = 3.6 km/hr × (5 mi/8 km) 
Result: miles per hr  1 m/s = 2.25 mi/hr 
Slow Driving Speed. Multiply both sides by 10,
10 m/s = 22.5 miles/hr
Fast Flying. Consider a more exciting speed, a rifle bullet has a muzzle velocity of about 1,000 feet per second = 1000 fps.What is that in miles per hour? ....km per hour?
Convert to mph  

1,000 ft per sec  = 1,000 fps (3,600 sec/1 hr) 
= 3,600,000'/hr (1 mi / 5,280')  
= 3,600/5.28 mi/hr  
= 682 miles/hr  
Convert to kph  
1000 fps  = 682 mi/hr (8 km / 5 mi) 
= 136 * 8 km/hr  
= 800 + 240 + 48 km/hr  
= 1,088 km/hr 
Convert fps to kps  

1000 fps  = 1,088 km/hr × 1hr/3600 sec 
= .302 km/sec 
ASTRONOMICAL UNIT
CONVENIENT MEASURE OF INTERPLANETARY DISTANCE.
CONVENIENT MEASURE OF INTERPLANETARY DISTANCE.
Astronomical unit (AU) is a unit of distance nearly equal to the radius of Earth's orbit around the Sun. The currently accepted value is 149,597,870 km (about 150 million kilometers or 93 million miles).
AU proves to be a convenient measure of distances within our Solar System. CONVENIENT MEASURE OF INTERSTELLAR DISTANCE. Interstellar distances require an even longer unit of distance. Commonly used is the lightyear or lightyear (symbol: ly), the distance light travels in a vacuum in one year. The term "light year" makes more sense if one recalls that velocity times time equals distance (see below). A lightyear equals 9,460,730,472,580.8 km (approximately 63,000 AUs). 
Lightspeed in a vacuum is an important physical constant denoted by the letter c (perhaps chosen from Latin word, celeritas: "swiftness"). "c" is the speed of all electromagnetic radiation, including visible light, in a vacuum. More generally, c is the speed of anything having zero rest mass. In metric units, c is exactly 299,792,458 metres per second (about 300,000 km/sec).
Recall that distance is speed × time.

If light can be thought of as speed, c,
and a year is a definite duration; then,,
and a year is a definite duration; then,,
"Light Year" is "c × year" = distance.
Granting the convenient approximation of c = 300,000 km/sec; then, how long does it take for a photon, particle of light, to travel from the Sun to the Earth?
Recall: time = distance divided by velocity (t = d /v)
t = 1 AU/c
t = 150,000,000 kms/300,000 kms/sec
t = 1,500/3 (sec) = 500 sec = 8 mins + 20 sec
Of course, it takes a year for that same photon to travel a light year (ly); thus, how many AUs in a LY?
We know it takes 500 seconds for a photon to travel one AU. Since it takes a year for a light particle (i.e., photon) to travel one Light Year (LY), we can determine AUs per LY by dividing 500 seconds into total seconds per one year.
We can easily determine total seconds per year with the expression:
1 year = 86,400 secs/dy × 365.25 days = 31,557,600 sec
To determine AUs traveled per year:
(31,557,600 sec/Yr ) × (AU / 500 sec) = 63,115.2 AUs/LY
For convenience, approximate l LY as 63,000 AU.
After a year, a photon travels one light year (LY), about a quarter of the distance to the nearest stellar system, Alpha Centauri.
ACCELERATION
VELOCITY CHANGE PER UNIT TIME.
Recall that velocity is the amount of distance traveled in a given period of time (distance per time). Any change in velocity requires acceleration; when decreasing velocity, one must decelerate (deceleration can be considered negative acceleration). We all experience acceleration daily when we travel from our home in our automobile; we must enter the vehicle when it's at rest (0 mph). After we strap in, the driver steps on the fuel peddle; and we accelerate (increase speed) to perhaps 60 mph. Similarly, when we reach our destination, the driver steps on the brake to decelerate (decrease speed) to 0 mph; then, we unstrap our safety restraints to exit the vehicle.
Newton stated that objects travel at a constant velocity until acted on by an external force. Virtually all scientists have long accepted this as true. This seems counter to our daily experience; for example, when our car is at 60 mph, we'll slow down as soon as the driver takes her foot off the fuel peddle (also know as "accelerator").
Does this prove Newton wrong?? Quite the contrary, he's proved correct; the vehicle will slow due to external forces such as air resistance and road friction.
In deep space (no air and no roads), once accelerated to 60 mph, we continue at that speed without consuming more fuel. (Of course, there'd be no stops for bathroom breaks as well as no breathing, but the fuel mileage would be excellent!!!)
Scientists, students and ordinary people have long observed that free falling objects do fall at 9.8 m/sec^{2}. (Note: feathers and parachutists are not really "free falling" because their downward progress is fortunately slowed by air resistance.)
For convenience, round 9.8 m/sec^{2} to 10 m/sec^{2}; this will facilitate further discussion on more thought experiments. That's to say, at 0 seconds, its downward velocity is 0 m/s; at 1 second, it's 10 m/s; at 2 secs, 20 m/s, and so on. Thus, velocity increases 10 m/s for every second of free fall.
Expressed as an equation: velocity = acceleration * time (v = a * t); if we're considering only the acceleration caused by gravity near Earth's surface, then we can substitute g for a.
v = g × t
Ten Seconds Acceleration
Accelerating at 10 m/sec^{2},
how fast are we after 10 sec??
Construct a simple table using arithmetic progression.
Simply increase speed 10 m/sec for every second elapsed.
ANSWER:
Start at rest (0 m/s) and accelerate at g; after 10 seconds, velocity is 100 m/s .
Equation at bottom of 2nd column:
velocity = time × g g = free fall acceleration due to gravity near Earth's surface. Thus,"gforce" acceleration. 

 Recall that 1 meter per second (m/sec) is 2.25 miles per hour (mph), a brisk walk.
 Ten times as fast (10 m/sec = 22.5 mph) is the speed of an automobile in slow traffic. A free falling object gains this much speed in one sec.
 In ten seconds, an free falling object's speed (100 m/sec) is 225 mph.
Following table reviews even higher speeds as well as a simple conversion technique. It's easier to appreciate meters per second if you know their equivalent speed in miles per hour.
Increasing Acceleration
In one minute, gforce acceleration can take us to over 1,000 mph. Only a few aircraft can achieve this speed.
To convert any velocity from meters per second to miles per hour (mph), just multiply by 2.25, relevant conversion constant.
After an hour of gforce acceleration, object can achieve asteroidal speeds, 36,000 m/sec = 36 km/sec. (Average asteroid speed is estimated as 25 km/sec.)

g =  9.8 m sec^{2}  =  9.8 m sec^{2}  ×  1 ft .3048 m  =  32.15 ft sec^{2} 
Since acceleration is velocity per time, also express as distance per time squared (i.e., m/sec^{2})?
Let a = v/t.
Recall that v = d/t; thus, a= (d/t) / t
a = d/t (1/t) = d/t^{2}
Acceleration equals distance divided by time squared.
Thus, acceleration is often used with units m/s^{2} or meters per seconds squared.
If a notional spaceship could achieve gforce acceleration, it would gain a final velocity of 864 km/sec after one day. 864 km/sec is about .005c (one half of one percent of light speed).
For this first day (or D_{1}) of gforce acceleration, the average velocity is 432 km/sec.
Ave V_{1} =  V_{Ini} + V_{Fin} 2  =  0 + 864 2  ×  km sec  =  432 km sec 
Thus, first day's total distance, d_{1}, is about 1/4 AU.
d_{1} =  432 km sec  ×  86,400 sec  =  37,324,800 km  ×  AU 149,597,870 km  ≈  1/4 AU 
d_{2} =  864 km sec  ×  86,400 sec  =  74,649,600 km  ×  AU 149,597,870 km  ≈  1/2 AU 
d_{3} =  d_{1}  ≈  1/4 AU 
Thus, a gforce vessel could conceivably travel 1 AU in three days.
d_{Ttl} =  d_{1 }+ d_{2 }+ d_{3}  ≈  .25 AU + .50 AU + .25 AU = 1.0 AU 
For more about gforce acceleration, see Thought Experiment's next chapter, "Accelerate for a Day".
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