Acceleration

[Simon]

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?

[tautech]

Don't ask me about the maths involved but IIRC from school, the mass of the accelerating object will be part of the calculation.

[Simon]

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.

[suicidaleggroll]

http://hyperphysics.phy-astr.gsu.edu/hbase/acons.html

y = 1/2 a * t^2

You know y and t, you can calculate a, and then v=a*t

In this case, yes, final speed is 2x average speed.

[madires]

a = (v2 - v1) / (t2 - t1)

[Simon]

a = (v2 - v1) / (t2 - t1)

Yes 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.

[IanB]

From my school physics lessons many eons ago...

For motion in a straight line with constant acceleration:

  v = u + at

  s = ut + ½at²

  s = ½(u + v)t

  v² = u² + 2as

where:

  u = initial velocity
  v = final velocity
  a = acceleration
  t = elapsed time
  s = distance travelled

[uncle_bob]

Hi

Simple 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

[IanB]

Acceleration is in meters per second squared. You continuously increase your velocity by some amount every second.

I find this much easier to grok if I say "acceleration is in meters per second, per second".

[Brumby]

From my school physics lessons many eons ago...

For motion in a straight line with constant acceleration:

  v = u + at

  s = ut + ½at²

  s = ½(u + v)t

  v² = u² + 2as

where:

  u = initial velocity
  v = final velocity
  a = acceleration
  t = elapsed time
  s = distance travelled

Oh - there's a real throw-back!!

... and yes, I still remember them.   (But we were only presented with 3.   s = ½(u + v)t  wasn't one of them.)