Question about physics/mechanics
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17-10-2012, 12:11 PM
Question about physics/mechanics
I've been studying a lot lately (hence my absence). I'm applying for a job as a freight train engineer. Because the company pays for the quite expensive training, they do an entrance exam. So I downloaded their syllabus booklet and read the thing. Most of the stuff i know from my days at technical school but some are buried deep behind the dust of time and the alcohol consumed back then. Blush

Anyway...
Here's my problem...
These formulas describe what they call the "uniformly accelerated motion" or "uniform acceleration"

[Image: formula.jpg]
Where:
s=distance traveled (meter)
a=Acceleration (meter/sec²)
Vo=original speed (meter/sec)
V=end speed (meter/sec)
t=time (sec)

my questions:
Where does the exponentiation of the time comes from in the denominator.
Why does it get divided by 2?

Some help would be appreciated...

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17-10-2012, 01:28 PM
RE: Question about physics/mechanics
(17-10-2012 12:11 PM)Observer Wrote:  I've been studying a lot lately (hence my absence). I'm applying for a job as a freight train engineer. Because the company pays for the quite expensive training, they do an entrance exam. So I downloaded their syllabus booklet and read the thing. Most of the stuff i know from my days at technical school but some are buried deep behind the dust of time and the alcohol consumed back then. Blush

Anyway...
Here's my problem...
These formulas describe what they call the "uniformly accelerated motion" or "uniform acceleration"

[Image: formula.jpg]
Where:
s=distance traveled (meter)
a=Acceleration (meter/sec²)
Vo=original speed (meter/sec)
V=end speed (meter/sec)
t=time (sec)

my questions:
Where does the exponentiation of the time comes from in the denominator.
Why does it get divided by 2?

Some help would be appreciated...

The t^2 is in the *numerator*. The bottom term is the denominator.,
"T" is squared because :
velocity is distance / time
accellertion is *change* in velocity / time.
that means accelleration i = (meters / second) / second, or meters/second^2
t^2 IS the "seconds squared"

The division by two comes from this :

distance, (or displaement) (ie your "s") = v(initial)*time + (1/2)*accceleration*t^2

Displacement equals the initial (or original) velocity multiplied by time plus one-half times the acceleration multiplied by time squared.
I've now confused myself trying to put this one into word...I'll let someone else say it.

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17-10-2012, 02:36 PM
RE: Question about physics/mechanics
(17-10-2012 01:28 PM)Bucky Ball Wrote:  The t^2 is in the *numerator*. The bottom term is the denominator.,
my bad
Quote:"T" is squared because :
velocity is distance / time
accellertion is *change* in velocity / time.
that means accelleration i = (meters / second) / second, or meters/second^2
t^2 IS the "seconds squared"

The division by two comes from this :

distance, (or displaement) (ie your "s") = v(initial)*time + (1/2)*accceleration*t^2

Displacement equals the initial (or original) velocity multiplied by time plus one-half times the acceleration multiplied by time squared.
I've now confused myself trying to put this one into word...I'll let someone else say it.
I get the "squared" part I guess Thumbsup

but le'ts wait for someone else to say it indeed Confused

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18-10-2012, 06:02 AM
RE: Question about physics/mechanics
Integration of x=1/2x^2+c. Velocity is the differential of acceleration. A constant rate (velocity) is distance divided by time, a constant acceleration is rate times time, resulting in a time squared. Thumbsup

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18-10-2012, 07:00 AM
RE: Question about physics/mechanics
Basically, it's easy if you have calculus.

Let's start with zero acceleration. That means *velocity is not changing*. Velocity is defined as rate of change of distance.

Let's call v_0 the constant velocity at time t = 0;

So with x_0 as your starting position at time t = 0 and x as your position at time t, we
can say that v_0 = (x - x_0) / t.

Or we can put it as
x = v_0 * t + x_0; // v_0 is a constant.

If you were to graph the velocity v on the vertical axis against time t on the horizontal, it would be a straight horizontal line.

Note that the *area under the graph* (between the horizontal line and the x-axis) is the distance travelled *from position x_0.

i.e. (x - x_0) = area under graph = v_0 * t ; // makes sense ?

Now. Let's consider the case where *the speed is changing at a constant rate a*. In essence, this is where the term in t^2 and the 1/2 comes from.

a = (v - v_0) / t.

If I graph velocity against time, I get a straight line, with positive or negative gradient (sloping upwards from left to right, or downwards from left to right respectively).

The key point is that *the area under the graph still gives the distance travelled from x_0* !!!

So now I have

x - x_0 = area under graph = Right angled triangle + a possible rectangle
= 1/2 base * height of triangle + base * height of rectangle
= 1/2 t * (a * t) + t * v_0
= 1/2 a * t^2 + v_0 * t;

This is about as simple as I can think of without calculus... based off some half remembered stuff also from my school days Big Grin It'd be more convincing with diagrams...
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19-10-2012, 12:22 AM
RE: Question about physics/mechanics
(18-10-2012 07:00 AM)morondog Wrote:  Basically, it's easy if you have calculus.

Let's start with zero acceleration. That means *velocity is not changing*. Velocity is defined as rate of change of distance.

Let's call v_0 the constant velocity at time t = 0;

So with x_0 as your starting position at time t = 0 and x as your position at time t, we
can say that v_0 = (x - x_0) / t.

Or we can put it as
x = v_0 * t + x_0; // v_0 is a constant.

If you were to graph the velocity v on the vertical axis against time t on the horizontal, it would be a straight horizontal line.

Note that the *area under the graph* (between the horizontal line and the x-axis) is the distance travelled *from position x_0.

i.e. (x - x_0) = area under graph = v_0 * t ; // makes sense ?

Now. Let's consider the case where *the speed is changing at a constant rate a*. In essence, this is where the term in t^2 and the 1/2 comes from.

a = (v - v_0) / t.

If I graph velocity against time, I get a straight line, with positive or negative gradient (sloping upwards from left to right, or downwards from left to right respectively).

The key point is that *the area under the graph still gives the distance travelled from x_0* !!!

So now I have

x - x_0 = area under graph = Right angled triangle + a possible rectangle
= 1/2 base * height of triangle + base * height of rectangle
= 1/2 t * (a * t) + t * v_0
= 1/2 a * t^2 + v_0 * t;

This is about as simple as I can think of without calculus... based off some half remembered stuff also from my school days Big Grin It'd be more convincing with diagrams...
I might be up to it....
I'll try and draw it it today and post it here to check if did I get it...

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19-10-2012, 12:46 PM
RE: Question about physics/mechanics
Hey morondog...
Here's my attempt to put it in a graph

[Image: 11206461.jpg]

The dashed and grey area under the graph is the distance travelled.
Is this correct?

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19-10-2012, 02:48 PM
RE: Question about physics/mechanics
Not sure about the distance if accel = infinity bit Tongue Otherwise yeah.

You can see, you plotted velocity against time. In this case a stationary object which then started moving with a constant positive acceleration.

To work out the distance travelled at constant speed
distance = time at constant speed x constant speed.

To work out the distance travelled after the constant acceleration started

distance = area of the triangle + area of the rectangle under the triangle
= 1/2 base x height of triangle + length x width of rectangle

the only hard bit maybe is realising that the height of the triangle is given by
height = acceleration x time at that acceleration.

By the way, you showed for positive slope - velocity increasing - but the formulae you derive work for negative slope (velocity decreasing) as well.

This is more of a visualization technique than a proof of anything Smile If you have any rudimentary calculus I can explain it in terms of calculus more easily.

Also note that if your acceleration varies - more often the case in the real world - then the distance travelled is still given by the area under the velocity-time graph.

Just remember, the term in t^2 comes from *constant acceleration*. If your acceleration is zero then it disappears, and you get the much more easily understandable formula
distance travelled from start = speed x time.

If on the other hand your acceleration is given by some other weird function, then your distance travelled formula will look a lot stranger and probably not have a t^2 term Wink But you don't need to worry about that. *However* interestingly enough, these are the exact principles which apply to for example, calculating orbits of planets, with some extra complications added to take into account that gravity (which produces acceleration) varies with distance, and that we're doing 2 or 3 d motion.

What I love about this stuff though, is that even with the simple formulae you can still calculate some useful results, depending on what you're applying them to.

Nice going man Smile
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19-10-2012, 11:13 PM
RE: Question about physics/mechanics
The visualisation helped perfect to made me understand what the formula tries to describe.

Btw: the accel=inifine part is just a description of the rectangle that gets divided.


My train won't accelerate that fast :-)

Thanks for helping me out guys.
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20-10-2012, 07:27 AM (This post was last modified: 21-10-2012 09:08 AM by TrulyX.)
RE: Question about physics/mechanics
You can think about it just using basic concepts that don't need an understanding of calculus, not actually having to understand integration and derivation. Some of the topics would be discussed in calculus or precalculus, but you don't need the background.

What the last equation is representing is a result of rate of change of position with respect to time (dS/dt) (derivative of position if you've had calculus), which is the velocity. (dS) just equals the change in position and is being divided by the change in time, (dt), which is by definition velocity: velocity is equal to displacement/time (e.g. meters/second or miles/hour).

One of the relationships you have to make, to get the motion equation you are referring to, is how to relate displacement. If you take the equation of velocity, displacement/time = velocity (d/t=v), and solve for displacement, (multiplying both sides), you get (d=v*t).

Another relationship you have to make is by taking into account that the velocity is you must use average velocity. The equation for average velocity is basically the average of the initial and final velocities [(vi+vf)/2]. Morondog kind of explained where that came from pretty well, and I really don't understand how to explain the equation any other way than realizing that relationship to position and the area under the curve/line (basically the basis for integral calculus) of that graph. It ends up being the mean, (how ever you want to look at it), and what Morondog explained.

Taking those things all together you get: (dS/dt) = [(vi+vf)/2].

Once you notice that you have yet to make the relationship between acceleration you can do that. You already have the equation (the first out of the two). It's the exact same relationship between velocity and position, just between acceleration and velocity. Acceleration is the rate of change of velocity, again that's done with respect to time (dv/dt= a). (Hence you get meters/seconds^2, because you are dividing velocity with respect to time, or position with time twice). Once you work that equation out, by multiplying by time, accounting for change in velocity (v broke down into vi and vf) and solving for the final velocity, you get that first equation (vf=at+vi).

Now, finally, you plug in for (vf) in [(vi+vf)/2] and solve.

(dS/dt) = [vi + (vi + at)]/2 -- That is where the divided by 2 comes from (average velocity).

Break down (dS/dt) (rate of change in position with respect to time) into [(sf-si)/t] (initial and final positions by time).

Doing that and a little algebra: [(sf-si)/t] = [(2vi+at)/2].

Multiply by (t) (solving for displacement) and you get: [s = (1/2)(2vi+at)(t)].

You get the squared when you distribute algebraically-- {s = (1/2)[(2vi)(t)+(at^2)]}

Then distribute back in the half (divide by two), cancel things out and you're done.

s= (at^2)/2 + vi(t)

Work back in the change position, optional, you'd get [sf = (at^2)/2 + vi(t) + si.]

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