*Notes to a video lecture on http://www.unizor.com*

__Constant Acceleration__

Let's consider a motion with constant

*vector of acceleration*from the position of Kinematics and derive simple equations that mathematically describe this motion in coordinate system.

Since

*acceleration*is the second derivative of a

*position*function and also the first derivative of a

*velocity*, the motion with constant

*vector of acceleration*

*{*

**A =****,**

*a***,**

*b***}**

*c**position*functions

**,**

*x(t)***and**

*y(t)***:**

*z(t)*

*x"(t) = a*

*y"(t) = b*

*z"(t) = c*where

**,**

*a***and**

*b***are constants**

*c*and at least one of these constants is not zero (otherwise, we deal

with a case of uniform motion considered in the previous lecture).

Let's represent this in vector form, considering that components of the

*position vector*

*are coordinate functions*

**P(t)****,**

*x(t)***and**

*y(t)***:**

*z(t)**{*

**P(t) =****,**

*x(t)***,**

*y(t)***}**

*z(t)**{*

**P'(t) =****,**

*x'(t)***,**

*y'(t)***}**

*z'(t)**{*

**P"(t) =****,**

*x"(t)***,**

*y"(t)***}**

*z"(t)*Therefore, in vector form equation of motion with constant

*acceleration vector*

**is:**

*A*

**P"(t) = A**The only function, whose derivative is constant, is a linear function. Hence, the first derivative of

*position*of our object, that is its

*velocity vector*, can be described by linear functions of time:

*x'(t) = a·t + v*_{x}(0)

*y'(t) = b·t + v*_{y}(0)

*z'(t) = c·t + v*_{z}(0)where

**,**

*v*_{x}(0)**and**

*v*_{y}(0)**are some constants.**

*v*_{z}(0)We have chosen the names of these constants to emphasize their meaning - components of

*velocity vector*

**at time**

*V(t)***:**

*t=0*

*x'(0) = v*_{x}(0)

*y'(0) = v*_{y}(0)

*z'(0) = v*_{z}(0)or, in vector form,

**= {**

*V(0)***,**

*v*_{x}(0)**.**

*v*_{y}(0)**}**

*v*_{z}(0)Continuing the vector representation of our equations of motion, we can express

*velocity vector*

**as a linear function of time using**

*V(t)**acceleration vector*

**and**

*A**initial velocity vector*

**as parameters:**

*V(0)***=**

*V(t)***·**

*A***+**

*t*

*V(0)*Now, since we have the first derivatives of

*position*functions, let's get the

*position*functions themselves.

The only function, whose derivative is a linear function, is a quadratic function. Hence, the

*position*functions can be described by quadratic functions of time:

*x(t) = a·t²/2 + v*_{x}(0)·t + x_{0}

*y(t) = b·t²/2 + v*_{y}(0)·t + y_{0}

*z(t) = c·t²/2 + v*_{z}(0)·t + z_{0}where

**,**

*x*_{0}**and**

*y*_{0}**are some constants equal to**

*z*_{0}*position*of our object at time

**:**

*t=0*

*x(0) = x*_{0}

*y(0) = y*_{0}

*z(0) = z*_{0}In vector form the above equations of motion can be represented shorter:

**=**

*P(t)***·**

*A***+**

*t²/2***·**

*V*_{0}**+**

*t***,**

*P*_{0}where

**=**

*V*_{0}**-**

*V(0)**initial velocity vector*and

**=**

*P*_{0}**-**

*P(0)**initial position*.

Obviously, the second derivatives of these three quadratic functions above correspondingly equal to

**,**

*a***and**

*b***- components of our constant**

*c**acceleration vector*of motion.

How to determine constants

**,**

*v*_{x}(0)**,**

*v*_{y}(0)**,**

*v*_{z}(0)**,**

*x*_{0}**and**

*y*_{0}**that participate in the equations for**

*z*_{0}*position*of our object?

Well, we cannot without additional information.

As we have determined,

**= {**

*V*_{0}**,**

*v*_{x}(0)**,**

*v*_{y}(0)**}**

*v*_{z}(0)*velocity vector*at the beginning of motion

**and**

*t=0***= {**

*P*_{0}**,**

*x*_{0}**,**

*y*_{0}**}**

*z*_{0}*initial velocity*and

*initial position*is needed to recreate the trajectory of an object moving with constant

*acceleration*in addition to the knowledge about its constant

*acceleration vector*

**= {**

*A***,**

*a***,**

*b***}**

*c*By itself, constant

*acceleration vector*does not define the motion, we need an

*initial position*and

*initial velocity*as well.

Knowing all of these, an

*initial position*, an

*initial velocity vector*and a constant

*vector of acceleration*, we can find equations of motion for

**,**

*x(t)***and**

*y(t)***.**

*z(t)*Does a motion with constant acceleration occurs along a straight line?

The answer depends on collinearity between an

*acceleration vector*and a vector of

*initial velocity*. The trajectory is a straight line if these two vectors are collinear, that is

*k·*

**a = v**_{x}(0)*k·*

**b = v**_{y}(0)*k·*

**c = v**_{z}(0)Or in vector form

*k·*

**A = V**_{0}In such a case

*x(t) = a·(t²/2 +**k*

**·t) + x**_{0}

*y(t) = b·(t²/2 +**k*

**·t) + y**_{0}

*z(t) = c·(t²/2 +**k*

**·t) + z**_{0}Or in vector form

*P(t) = A·(t²/2 +**k*

**·t) + P**_{0}The equations above define a trajectory collinear to vector

**= {**

*A***,**

*a***,**

*b***}**

*c**uniform motion*.

If the motion with a constant

*acceleration*occurs within a

straight line (like in case of a car, accelerating along a straight

road), it's very convenient to choose a system of coordinates, where

X-axis coincides with this line and the positive direction of the X-axis

coincides with the direction of the motion.

Then

**and**

*y(t)=z(t)=0***for all moments of time**

*y'(t)=z'(t)=0***. In particular,**

*t**velocity vector*looks now as

**,**

*x'(t)***,**

*0***}**

*0**acceleration vector*looks like

**,**

*a***,**

*0***}.**

*0*For these cases traditionally the letter

**is used for the X-component of the**

*v**velocity vector*. So, we will use in these cases the symbol

**for X-component of the**

*v(t)**velocity vector*

**.**

*x'(t)*The only non-trivial equation of motion then will be about

**:**

*x(t)*

*x(t) = a·t²/2 + v*_{0}·t + x_{0}(where

**is**

*x*_{0}=x(0)*initial position*,

**is**

*v*_{0}=v(0)*initial speed*of an object at the start of motion and

**- its**

*a**constant acceleration*).

If the origin of the system of coordinates coincides with the initial position of the object,

**and the equation of motion looks simpler:**

*x*_{0}=0

*x(t) = a·t²/2 + v*_{0}·tIf, in addition, the object is at rest in the beginning of motion,

**and the equation of motion looks really simple:**

*v*_{0}=0

*x(t) = a·t²/2*Since the positive direction of the X-axis is the same as the direction of the movement,

*velocity vector's*direction plays minimal role, and only its magnitude (positive) is important, in which case we can use the term

*speed*for

**instead of**

*v(t)**velocity*.

CONCLUSION

An object moving from the

*initial position*

*{*

**P**_{0}=**,**

*x*_{0}**,**

*y*_{0}**}**

*z*_{0}*velocity vector*

*{*

**V**_{0}=**,**

*v*_{x}(0)**,**

*v*_{y}(0)**}**

*v*_{z}(0)*acceleration vector*

*{*

**A =****,**

*a***,**

*b***}**

*c*

*x(t) = a·t²/2 + v*_{x}(0)·t + x_{0}

*y(t) = b·t²/2 + v*_{y}(0)·t + y_{0}

*z(t) = c·t²/2 + v*_{z}(0)·t + z_{0}Or, in vector form,

**=**

*P(t)***·**

*A***+**

*t²/2***·**

*V*_{0}**+**

*t*

*P*_{0}If the

*initial position*of an object is on the X-axis,

*initial velocity vector*and

*acceleration vector*are colinear with the X-axis, the equations are simplified:

*x(t) = a·t²/2 + v*_{0}·t + x_{0}

*y(t) = 0*

*z(t) = 0*If, in addition, the origin of coordinates coincides with the point of beginning of motion, that is

**, the equations are simplified even further:**

*x*_{0}=0

*x(t) = a·t²/2 + v*_{0}·t

*y(t) = 0*

*z(t) = 0*Finally, if, in addition, the starting velocity is zero, that is

**, the equations are simplified even further:**

*v*_{0}=0

*x(t) = a·t²/2*

*y(t) = 0*

*z(t) = 0*
## No comments:

Post a Comment