Explain why 60mph is the same as 88ft/s?
112479
60 miles      5280 ft       1 hr           1 min\n---------  x --------   x ----------  x ---------  =\n1 hour          1 mi        60 min         60 sec\n\n60(5280) ft (because all other units cancel)\n-------------------------------------------\n60(60) sec (because all other units cancel)  Its simple \n\nwhen you use unit multipliers.
412765
1m/s physically you are walking towards a stationary object that is reflecing your image, as such you are walking towards it at the steady 1m/s.\n\nIf not that then you are both approaching a common point at 1 m/s so you combine your velocities to get 2 m/s.  but that's semantic bullcrap\n\nYou could get really technical and add in the fact that the photons are travelling back at you at the speed of light but that's just stupid.
652370
Okay, I've thought about this and it's not possible to determine the rate of acceleration from the given criteria.  It is an "average" of 60 miles per second, BUT how many miles in the first second compared to the number of miles in the final, fifth second?  If that is known, you can project MPH from seconds 6 to 10 or the additional 1/4 mile.  How do you calculate the rate of acceleration in your theory?  How fast is the vehicle traveling at each point of the 1/4 mile distance?
392981
Is the speed at which it travels.
758481
It's 8 meters per second.  He's gone 5000 meters in 625 seconds; 5000 meters/625 seconds = 8 meters per second.\n\n v=d/t\nv=5000 m/625 seconds\nv=8m/sec
554079
(150mi/1h)*(1h/60min)*(1min/60sec)*(5280ft/1mile)=220ft/sec\n\nFirst, convert miles to feet: 150 miles * 5280 ft/mile = 792,000.  This is your feet/hour.\n\nNow, convert hours to seconds: 1 hr * 60 min/hr * 60 sec/min = 3600 seconds in one hour.\n\nDivide your feet/hour by seconds/hour:\n\n792,000/3600 = 220 ft/sec
931765
1. d=t, any form of d=t+b would be parallel. b = any number. That is a line.\n\n2. d=t would also work here.\n\n3. Well if you go 60 MPH and you wnat to know how many miles a minute you can go convert using (60miles/hour)*(1hour/60minutes) and you get 1mile/minute. If you need feet, 1 mile is 5280 feet, so 5280 feet/minute
125621
B) Increase\n\nThe difference in the two objects' velocities would be constant since that are both accelerating at the same rate, but the first object already had a higher initial velocity (from Object B' point of view).\n\nThe equation relating the distance traveled with respect to initial velocity, accelertion and time is,\nd = v_i * t + 1/2 a*t^2\nSo for both objects, after B is released the 1/2 a*t^2 part of the equation is the same, so we can ignore that part and focus on the v_i * t part.\nIf we start t=0 at the moment object B is released, object A's velocity is some non-zero number while B has no initial velocity.  So as time goes on, the distance between the 2 objects inceases.
979735
If we just give you the answer, you'll never learn properly.\n\nAs has been suggested you need to get everything in the right units first!!\n\nNewtons 2nd law F= Ma\n\nForce = Mass X Acceleration\nForce is in Newtons\nMass in Kilograms\nAcceleration in Metres per second\n\nYou already have the mass - 38,000kg\n\n\nYou need to figure out the acceleration:\n\n\n60 kilometres/hr = 60,000metres/hr\n60,000metres/hr divided by 60 = 1,000 meters per minute\n1,000 metres per minute divided by 60 will give you the needed finishing velocity of metres per second\n\nAcceleration = finishing velocity / time ( 20 seconds )\n\nForce therefore = 38,000 X your answer for acceleration!
378965
975 m/s (3,200 ft/s), 884 m/s (2,900 ft/s) effective to 600 yards. 5.56mm
806168
Wait, so you are saying he is moving 60mph, and he throws 60mph in the direction he was coming from?  If you sum the velocities you get zero; the ball would drop straight down (from your point of view).  To him, it would appear to be moving at 60mph.\n\nIf he simply dropped the ball, it would be moving at 60mph alongside him.  To him, it would look like it were falling straight down.  You can test this; toss a ball straight up in a moving vehicle - as long as you do not turn a corner or change speed, the ball will move straight up and come back down.\n\nObviously wind resistance at 60mph is significant, but since you said it was in a vacuum, there is no wind.  If you tried tossing a ball up in a convertible with the roof open, it would decelerate a bit.
120276
The first thing you have to understand is that the speed of light in a vaccum is constant, no matter how fast you're moving, nor how fast the source of the light is moving. This is bizarre. If someone standing on a conver tosses a ball to you at 5 mph while you're standing on the corner, you'll see the speed of the ball as being 5 mph relative to you. But if you're driving at 15 mph and they throw it at you at 5 mph, the speed of the ball relative to you will be 20 mph. This is how most things work. But with light, no matter the speed of the source or yourself, you'll always measure the speed of light as being 299,792,458 metres per second.\n\nEinstein accounted for this by postulating his special theory of relativity, which says basically that the reason for this is because if you move really fast, three things happen: 1) time slows down, 2) mass increases, and 3) length in directions parallel to the direction of movement contract. If you take those three hypotheses, you'll also note that the concept of things being "simulatanous" is a relative term, meaning two observers moving at the same velocity (or moving at no speed relative to one another) will agree on which events are simultaneous and which are not. However, observers at different velocities will have different concepts of what is and isn't simulataneous--and yet THEY'RE BOTH CORRECT. It's pretty mind-blowing, but if you take all of these notions together, you get a consistent view of reality.\n\nThat's Einstien's SPecial Theory of Relativity. About a decade later he formed his far more complicated general theory of relativity.\n\nThe general theory says that Newton's view that gravity operates by exerting a force on a falling object is incorrect. Rather, by general relativty, a free-falling object that is in a vacuum has no forces acting on it at all. The movement of falling objects isn't due to force, but rather the curvature of space-time--massive objects actually curve the surrounding universe. In Newtonian physics, we regard there being three spacial dimensions (up/down, left/right, and backwards/forwards) and a single dimension of time. In general relativity, there are four dimensions of space-time. In Newtonian physics, a foce makes the earth orbit around the sun. In general relativity, the earth and the sun curve the surrounding space-time continuum, and both bodies move in a straight line through curved space, resulting in the perceived sense that one body is circling about the other. In Newtonian physics, light should not bend to gravity, because gravity affects massive bodies, and light has no mass. But in fact, light does bend around stars and such.\n\nGeneral relativity has been likened to a stretched sheet of elastic material, like a trampoline. Heavier objects sink more, and make the sheet curve about them, and light obejcts rolling across are deflected by this curving.\n\nGood luck understanding all that. Hope your head doesn't explode.
128774
Obviously, the 700 would be quicker. I can't give you exact numbers tho. I can tell you that 60mph on a quad feels like 120mph, and they are not designed to be ridden that fast.
1184449
It does not matter how many seconds your car moves in first gear, after that it should be 60 miles/hour and the feets you do the math by yourself. GOT IT???DUHHH
403577
You have miles per minutes, and you want to convert it to miles per hour.  Set it up as fractions so the units cancel out, like this:\n\n1mile/8minutes * 60minutes/1hour  Now the minutes cancel and miles/hour is the final units.  Now just do the math.  60/8 = 7.5 mph.
426864
Roughly 3 ten-thousandths of a second.\n\nIf an object is travelling 60 mph, it moves at 1 mile or 5,280 ft/min.  60 seconds in an minute 5,280/60 = 88 ft per second.  12 inches in a foot = 1,056 inches per second.  2.76/1056 = approx. .00261 seconds.
1099642
You should probably skip what I wrote and go straight to the link http://whatis.techtarget.com/definition/0,,sid9_gci529957,00.html :-)\n\nYou measure the distance fallen in meters.  Let us say that this distance as a function of time t in seconds since the ball was dropped in a vacuum is s(t) meters.\n\nThe "instantaneous" downward velocity of the ball at time t is just the derivative of s(t) with respect to t ... for example in the h seconds between time t = k and time t = k + h, the ball drops an extra s(k + h) - s(k) meters ... so the average velocity between times t = k and t = k + h is (s(k + h) - s(k)) / h, and the limit as h -> 0 of (s(k + h) - s(k)) / h is the velocity at time t = k and is the derivative of s(t) at t = k, and you can see that it has units meters per second because s(k+h) - s(k) is in meters and h is in seconds.\n\nSo distance is meters and velocity is meters per second, with v(t) = s'(t) the derivative of s with respect to t.\n\nNow acceleration is defined as the rate of change with repsect to time of velocity.  You will have a(t) = v'(t) = s''(t), by the same reasoning as before.  The velocity at time t = k is v(k) = s'(k) and the velocity at time t = k + h seconds is v(k + h) = s'(k + h).  The average acceleration between times t = k and t = k + h is then (v(k + h) - v(k)) / h in units of (meters per second) per second, or meters per second squared.  The instantaneous acceleration at time t = k is the limit of that as h -> 0 and is v'(t) = s''(t) at t = k.\n\nFalling in a vacuum at constant acceleration g = 9.8 meters per second squared means you have\n\nacceleration = constant g = 9.8 meters per second squared\n\nincreased downward velocity after falling t seconds = g t = (9.8 meters per second squared) time (t seconds) = 9.8 t meters per second.\n\ndistance fallen (if dropped with initial velocity zero) after falling t seconds is (1/2) g t^2 = (1/2) (9.8 meters per second squared) (t seconds)^2 = (1/2) g t^2 meters.\n\nThe rate of change of acceleration would be meters per second cubed.