Acceleration = (change in speed) divided by (time for the change)
Acceleration = (25 - 0)/(3) = 81/3 miles per second2
Don't try this at home. That acceleration is about 1,368 G's. You can not survive it!
Average acceleration = (change in speed) / (time for the change) = (60 - 0) / (2.1) = 28.571 miles per hour per second, or 41.905 feet per second2 Using: (1 mile / hour - second) x (5,280- ft / mile) / (3,600 second / hour) = 1.4666 ft/sec2
0.03 seconds.
Standard form: 1.5*10-1 seconds, expanded form: (0.1 + 0.05) seconds.
For a rough estimate, travelling for 13 hours would be 845 miles (13*65=845)A fairly exact solution:844 miles/65mph=12.98 hours.0.98 hours = 58.8 minutes0.8 minutes = 48 seconds12 hours 58 minutes 48 secondsFor an exact solution:Every hour has 60 minutes or 3600 seconds. 65 miles per hour is a mile every 3600/65 seconds.844 miles * 3600/65 seconds = 3038400/653038400/65 simplified is 607680/13.607680/13 is exactly the number of seconds it would take. That is 46744.615384... seconds. The 615384 repeats endlessly.46744.615384... seconds is equal to 12 hours 59 minutes 4 seconds .615384... seconds. 607680/13 is the answer in fraction form.
Time = distance/speed = 16/45 = 0.35556 hours that is the answer in decimal form. Now if you wish you can convert decimal time to hours and minutes: 0.35556 hours = 0.35556 * 60 = 21.3333 minutes 0.3333 minutes = 0.3333 *60 = 20 seconds so the answer in this form is: 21 minutes and 20 second
Average acceleration = (change in speed) / (time for the change) = (60 - 0) / (2.1) = 28.571 miles per hour per second, or 41.905 feet per second2 Using: (1 mile / hour - second) x (5,280- ft / mile) / (3,600 second / hour) = 1.4666 ft/sec2
0.5 seconds
1 billionth of a second in standard form is: 1.0 × 10-9 seconds.
2nd:Law Of Acceleration.3rd:Law Of Action/Reaction.
0.03 seconds.
Since the derivative of velocity is acceleration, the answer would be technically 'no'. Here is why: v = 0 v' = 0 = a Or in variable form... v(x) = x v(0) = 0 v'(0) = 0 = a You can "trick" the derivative into saying that v'(x) = 1 = a (since the derivative of x = 1) and then stating v'(0) = 1 = a... but that is not entirely correct. Acceleration is a change over time and is measured at more then one point (i.e. the acceleration of this body of matter is y from time 1 to 5) unless using derivatives to form the equation of the acceleration line/curve. If an object has a constant acceleration of 1, then the velocity is constantly increasing over that time. Using the equation discussed above and looking at acceleration over time, at 0 seconds, acceleration is 0 and so is velocity, but from 0-1 seconds acceleration is 1 and velocity is 1 as well. 0-2 seconds, acceleration is 1, but velocity would be 2 (at the end of 2 seconds).
It is 1.5*10-1 seconds, which is 10 + 5 seconds.
There are several formulae that involve acceleration. The most basic one is the definition of acceleration, which is: a = (difference in velocity) / time This assumes constant acceleration. For non-constant acceleration, the more general formula is: a = dv / dt where "dv" is the difference in velocity, and "dt" is the time interval, with the additional assumption that it is a very small time interval. For more details, read an introductory calculus book, to understand the concept of "derivative".
Gravity is a form of acceleration and so is measured in metres per second^2.
Gravity is a form of acceleration and so is measured in metres per second^2.
Standard form: 1.5*10-1 seconds, expanded form: (0.1 + 0.05) seconds.
1.0 × 10-9 seconds.