Acceleration - Page 1
Forum section: electronics / general

author: Simon
author_rank: Unknown
timestamp: July 07, 2016, 07:42:14 pm
content: I'm doing my assignment for one of my HNC modules. I'm being asked to draw an acceleration diagram for something that has a uniform acceleration so yes something really boring like a slope. The module has not actually taught me how to work out the speed achieved. I am told that an object accelerates from 0 uniformly and travels a certain distance in a certain amount of time. So I can work out the average speed but I don't actually know the acceleration. I will hazard a guess that this lies in calculus. It would be nice to put an actual value on the top of my acceleration diagram showing that the object starts at 0 and ends at a certain speed. Would I be correct introducing because I'm struggling to find any formulas and explanations for this online that knowing the average speed of something that uniformly accelerates I can assume that the final speed is twice the average speed?

author: tautech
author_rank: Super Contributor
timestamp: July 07, 2016, 08:00:58 pm
content: Don't ask me about the maths involved but IIRC from school, the mass of the accelerating object will be part of the calculation.

author: Simon
author_rank: Unknown
timestamp: July 07, 2016, 08:02:32 pm
content: no not in this, they just say that it moves from a standstill to 5m in 0.92s so I have the average speed, I'd assume the final speed is that x2 for linear acceleration.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: July 07, 2016, 08:10:13 pm
content: http://hyperphysics.phy-astr.gsu.edu/hbase/acons.htmly = 1/2 a * t^2You know y and t, you can calculate a, and then v=a*tIn this case, yes, final speed is 2x average speed.

author: madires
author_rank: Super Contributor
timestamp: July 07, 2016, 08:18:31 pm
content: a = (v2 - v1) / (t2 - t1)

author: Simon
author_rank: Unknown
timestamp: July 07, 2016, 08:24:52 pm
content: Quote from: madires on July 07, 2016, 08:18:31 pmYes i think I worked it out, according to the course notes I calculate the average speed , then I can calculate the acceleration and therefore the speed I'd get to at any given time.

author: IanB
author_rank: Super Contributor
timestamp: July 07, 2016, 08:34:23 pm
content: From my school physics lessons many eons ago...For motion in a straight line with constant acceleration:v = u + ats = ut + ½at²s = ½(u + v)tv² = u² + 2aswhere:u = initial velocityv = final velocitya = accelerationt = elapsed times = distance travelled

author: uncle_bob
author_rank: Unknown
timestamp: July 07, 2016, 08:48:06 pm
content: HiSimple way to look at it:Velocity is in meters per second. You have an average speed of 10 meters a second.Acceleration is in meters per second squared. You continuously increase your velocity by some amount every second. If you have an acceleration of 1 meter per second squared, at about 10 seconds, it gets you to 10 meters a second.Yes there is an *implied* step process in the stuff above. It's not really there. You do not instantly step the velocity at the end of each second....Bob

author: IanB
author_rank: Super Contributor
timestamp: July 07, 2016, 08:59:44 pm
content: Quote from: uncle_bob on July 07, 2016, 08:48:06 pmI find this much easier to grok if I say "acceleration is in meters per second, per second".

author: Brumby
author_rank: Unknown
timestamp: July 08, 2016, 08:19:56 am
content: Quote from: IanB on July 07, 2016, 08:34:23 pmOh - there's a real throw-back!!... and yes, I still remember them. (But we were only presented with 3.s = ½(u + v)twasn't one of them.)