calculas for phyiscs

constant acceleration

Calculus is an advanced math topic, but it makes deriving two of the three equations of motion much simpler. By definition, acceleration is the first derivative of velocity with respect to time. Take the operation in that definition and reverse it. Instead of differentiating velocity to find acceleration, integrate acceleration to find velocity. This gives us the velocity-time equation. If we assume acceleration is constant, we get the so-called first equation of motion [1].

a	=	dv dt
dv	=	a dt
v ⌠ ⌡ dv v0	=	t ⌠ ⌡ a dt 0
v − v0	=	at
v	=	v0 + at [1]

Again by definition, velocity is the first derivative of position with respect to time. Reverse this operation. Instead of differentiating position to find velocity, integrate velocity to find position. This gives us the position-time equation for constant acceleration, also known as the second equation of motion [2].

v		=	ds dt
	ds	=	v dt
	ds	=	(v0 + at) dt
s ⌠ ⌡ ds s0		=	t ⌠ ⌡ (v0 + at) dt 0
s − s0		=	v0t + ½at2
s		=	s0 + v0t + ½at2 [2]

Unlike the first and second equations of motion, there is no obvious way to derive the third equation of motion (the one that relates velocity to position) using calculus. We can't just reverse engineer it from a definition. We need to play a rather sophisticated trick.

The first equation of motion relates velocity to time. We essentially derived it from this derivative…

dv	= a
dt

The second equation of motion relates position to time. It came from this derivative…

ds	= v
dt

The third equation of motion relates velocity to position. By logical extension, it should come from a derivative that looks like this…

dv	= ?
ds

But what does this equal? Well nothing by definition, but like all quantities it does equal itself. It also equals itself multiplied by 1. We'll use a special version of 1 (^dt_dt) and a special version of algebra (algebra with infinitesimals). Look what happens when we do this. We get one derivative equal to acceleration (^dv_dt) and another derivative equal to the inverse of velocity (^dt_ds).

dv	=	dv		1
ds		ds
dv	=	dv		dt
ds		ds		dt
dv	=	dv		dt
ds		dt		ds
dv	=	a		1
ds				v

Next step, separation of variables. Get things that are similar together and integrate them. Here's what we get when acceleration is constant…

dv ds		=	a  1 v
v dv		=	a ds
v ⌠ ⌡ v dv v0		=	s ⌠ ⌡ a ds s0
½(v2 − v02)		=	a(s − s0)
	v2	=	v02 + 2a(s − s0) [3]

Certainly a clever solution, and it wasn't all that more difficult than the first two derivations. However, it really only worked because acceleration was constant — constant in time and constant in space. If acceleration varied in any way, this method would be uncomfortably difficult. We'd be back to using algebra just to save our sanity. Not that there's anything wrong with that. Algebra works and sanity is worth saving.

v =	v0 + at	[1]
		+
s =	s0 + v0t + ½at2	[2]
		=
v2 =	v02 + 2a(s − s0)	[3]

constant jerk

The method shown above works even when acceleration isn't constant. Let's apply it to a situation with an unusual name — constant jerk. No lie, that's what it's called. Jerk is the rate of change of acceleration with time.

j =	da
	dt

This makes jerk the first derivative of acceleration, the second derivative of velocity, and the third derivative of displacement.

j =	da	=	d2v	=	d3s
	dt		dt2		dt3

The SI unit of jerk is the meter per second cubed.

⎡ ⎢ ⎣	m/s3	=	m/s2	⎤ ⎥ ⎦
			s

An alternate unit is the g per second.

⎡ ⎢ ⎣	g	=	9.80665 m/s2	= 9.80665 m/s3	⎤ ⎥ ⎦
	s		s

Jerk is not just some wise ass physicists response to the question, "Oh yeah, so what do you call the thirdderivative of displacement?" Jerk is a meaningful quantity.

Jerk is the derivative of acceleration. Undo that process. Integrate jerk to get acceleration as a function of time. I propose we call this the zeroeth equation of motion for constant jerk. The reason why will be apparent after we finish the next derivation.

j =	da
	dt
da =	j dt

a		t
⌠ ⌡	da =	⌠ ⌡	j dt
a0		0

a − a0 =	jt
a =	a0 + jt	[0]

Acceleration is the derivative of velocity. Integrate acceleration to get velocity as a function of time. We've done this process before. We called the result the velocity-time relationship or the first equation of motion when acceleration was constant. We should give it a similar name. This is the first equation of motion for constant jerk.

a =	dv dt
dv =	a dt
dv =	(a0 + jt) dt

v		t
⌠ ⌡	dv =	⌠ ⌡	(a0 + jt) dt
v0		0

v − v0 =	a0t + ½jt2
v =	v0 + a0t + ½jt2	[1]

Velocity is the derivative of displacement. Integrate velocity to get displacement as a function of time. We've done this before too. The resulting displacement-time relationship will be our second equation of motion for constant jerk.

v =	ds dt
ds =	v dt
ds =	(v0 + a0t + ½jt2) dt

s		t
⌠ ⌡	ds =	⌠ ⌡	(v0 + a0t + ½jt2) dt
s0		0

s − s0 =	v0t + ½a0t2 + ⅙jt3
s =	s0 + v0t + ½a0t2 + ⅙jt3	[2]

Please notice something about these equations. When jerk is zero, they all revert back to the equations of motion for constant acceleration. Zero jerk means constant acceleration, so all is right with the world we've created. (I never said constant acceleration was realistic. Constant jerk is equally mythical. In hypertextbook world, however, all things are possible.)

Where do we go next? Should we work on a velocity-displacement relationship (the third equation of motion for constant jerk)?

v =	v0 + a0t + ½jt2	[1]
		+
s =	s0 + v0t + ½a0t2 + ⅙jt3	[2]
		=
v =	f(s)	[3]

How about an acceleration-displacement relationship (the fourth equation of motion for constant jerk)?

a =	a0 + jt	[1]
		+
s =	s0 + v0t + ½a0t2 + ⅙jt3	[2]
		=
a =	f(s)	[4]

• The derivative of position with time is velocity (v = ^ds_dt).
• The derivative of velocity with time is acceleration (a = ^dv_dt).

or integration (finding the integral)…

• The integral of acceleration over time is change in velocity (∆v = ∫a dt).
• The integral of velocity over time is change in position (∆s = ∫v dt).

Here's the way it works. Some characteristic of the motion of an object is described by a function. Can you find the derivative of that function? That gives you another characteristic of the motion (or maybe it gives you something only a mathematician would love). Can you find its integral? That gives you a different characteristic (or maybe a load of nonsense that only a mathematician would love). Repeat either operation as many times as necessary (probably no more than twice). Then apply the techniques and concepts you learned in calculus and related branches of mathematics to extract more meaning — range, domain, limit, asymptote, minimum, maximum, extremum, concavity, inflection, analytical, numerical, exact, approximate, and so on. I've added some important notes on this to the summary for this topic.