The time it takes a body in motion moving from point A to point B ,is benchmarked with a standard of time,?
462967
yes, space and time are forever linked into spacetime. please read the pages to which the following link leads. i hope they answer your question.
35172
Can be any measurement you want. V will end up being correlated, i.e., miles per hour, feet per second, inches per days, etc. \n\nIn V=5280 feet/60 minutes, V would equal feet per minute. You can then convert that to whatever. In the example, V=88 feet per minute, 1 mile per hour, etc.
815662
At one number per second, it would take you 11, 570,000 days, or about 31709 years.\n\neven at 10 per second you would need 3170 years, so, being a mere human, you couldn't do it.\n\nWill
265901
average speed: total distance/total time\ntotal distance: 60 miles\ntotal time: 100 minutes = 1,6666666666666667 hour\naverage speed = 60 miles/1,6666666666666667 hour\naverage speed = 36 mph\n---------------------------------------------------\nvelocity is zero because man does not change of position
202957
How far apart are the marks?\nTime taken to travel between the marks?\nSpeed = distance divided by time.
9180
This would depend on how fast you count.  Even a small variance in the amount of time it takes to say each number would have a HUGE impact when accumulated over 1 billion counts.  So we can only make an "If, then" statement... That is "If we count two numbers per second, then this is how long it would take..."\n\nSo let's say you counted exactly two numbers per second.  It would take you 2 billion seconds to do this.  That's somewhere around 63 and a half years, if I did the math right.
765042
D = distance in feet\nt = time in seconds\nS = speed in miles per hour\n\n[D feet x (1 mile/5280 feet)] / [t seconds x (1 hour/3600 seconds)]\n= S miles per hour\n\nDoing the arithmetic, we get:\n0.6818 D / t = S mph\n\nIf D is 150 feet and t is 4 seconds, \n0.6818 (150)/ 4 = 25.56 mph\n\nIf the time is 8 seconds, the speed would be half as much, or \n12.78 mph
900973
At noon\nA = (0, 0), B = (-90, 0)\n\nAt time t hours\nA = (0, 40t) B = (-90 + 60t, 0)\n\nSo |AB| = √[(40t)² + (-90 + 60t)²]\n= 10√[16t² + 81 - 108t + 36t²]\n= 10√[52t² - 108t + 81]\nd|AB|/dt = 5(104t - 108)/√[52t² - 108t + 81]\n\nWhen t = 1 d|AB|/dt = 5 * (-4)/√[52 - 108 + 81]\n= -20/√25\n= -4 mph ie they are getting closer at a speed of 4 mph
927736
Everyone above assumed a constant speed.  the more general formula is \n\nD = ∫v dt
748368
first get the time from the horizontal motion  t = .51 / 1.2 s (uniform motion, no acceleration here)\n\nthen use h = 1/2gt^2 for the vertical motion (with initial velocity zero)
208867
Time is a point of "relevance". It is measured only by what you measure it by. a month is a month of 30/31 days measured by calender, or new moon to new moon measured by moon phases.  In the case of an air plane, if the plane flies for an hour in any direction, it is an hour later than when you took off. In this example lets say it takes 4 hours to fly from new york to los angeles. there are four time zones involves so there is a three hour difference. You take off at 12 noon from new york. You fly 4 hours to LA. It is now 4 pm (new york time or airplane time) Since it was 9 am (three hours differenct from new york) los angeles time when you took off, you flew for four hours it is now 1 pm in LA. You still flew for FOUR hours,\n\nThere is no relevance between time and gravity in a traveling situation. Our moving speed is relative to the earth. 2 people walking together, to each other no speed difference.  One stand s still he will perceive the speed of the other.  All movement ideas suggested in these answers are correct.  "No gravity" can even be experienced even when in the atmosphere or even on earth, again no relation to the speed question at hand. A vehicle moves (speed) in relation to a point on earth, regardless if the earth is moving, (we are moving with the earth so that is a null point). Mathmatical: x = speed of earth  Y = speed of plane t = time d = distance s= starting point (given directions are opposite)\n\ns + (y x t) = d           d - (y x t) = s\n\nNote for the avionic people, planes have air speed indicators which give a different value than a speedometer
45760
Time is currently one of the few fundamental quantities (quantities which cannot be defined via other quantities because there is nothing more fundamental known at present). Thus, similar to definition of other fundamental quantities (like space and mass), time is defined via measurement. Currently, the standard time interval (called conventional second, or simply second) is defined as 9 192 631 770 oscillations of a hyperfine transition in the 133Cs atom.
286517
It differs. From 90 feet a minute to 180 feet a minute.
630244
No not at all. This concept has been tested on more than one occasion. The faster a body moves the slower time passes relative to that person. This was tested by getting two superbly accurate atomic clocks were taken. One was left on earth and another was taken up in a jet plane. When the plane landed the clock on the plane was observed to be a couple fractions of a second behind the clock on earth. \n\nThe equation for this difference is:\n\ndelta T(stationary)=(1/sqrt(1-v^2/c^2))*delta T(moving)\n\nNot to be disrespectful but why are most of your questions concerned with doubting the theories that Einstein has put forth. Almost all of his theories have been tested and confirmed to extremely high degrees of accuracy
465183
Earth's speed changes from time to time.  The average is 29.783 kilometers/second or 107,218 kilometers/hour.
687277
1 hour or 60 mins or 3600 seconds :-) is the time it takes but it will still be 12:30 :-p
655000
The minute hand moves at one rate, the hour hand at another, so we first need to determine the rate (in degrees of the rotation) that each travels.\n\nEach hour, the minute hand moves 360 degrees, and the hour hand moves only 1/12 of that:  360 * 1/12 = 30 degrees.  Each minute, the minute hand moves 1/60 of the entire revolution: 360 * 1/60 = 6 degrees.  The hour hand, over the course of one minute would again move 1/12 of what the minute hand moved: 6 * 1/12 = 0.5 degrees.  \n\nNow, let's consider the starting point of each hand.  The minute hand starts at the 12:00 position, which we'll refer to as 0 degrees.  The hour hand starts at the 5:00 position, which would be 5/12 of the total rotation (360 degrees) or:  5/12 * 360 = 1800/12 = 150 degrees.\n\nSo, the position of the minute hand at any given time (t) in minutes would be represented by the starting point (0 degrees) plus 6 degrees times the number of minutes (t) or:\n\n0 + 6t    or    6t\n\nThe location of the hour hand will be represented by the starting point (150 degrees) plus 0.5 degrees times the number of minutes (t) or:\n\n150 + 0.5t\n\nThen, find the point where these two locations are equal:\n\n6t = 150 + 0.5t\n\nsubtract 0.5t from each side:\n\n6t - 0.5t = 150 + 0.5t - 0.5t\n5.5t = 150\n\ndivide both sides by 5.5:\n\n5.5t/5.5 = 150/5.5\nt = 27.27\n\nSo, the two hands will be together at 27.27 minutes past 5:00.