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

__Acceleration__

Recall that, if

**,**

*x(t)***and**

*y(t)***are coordinate functions of the moving object, the derivatives of these functions**

*z(t)***,**

*x'(t)***and**

*y'(t)***represent components of a velocity of this moving object, that is a vector of instantaneous change of an object's position.**

*z'(t)*Let's generalize. If function

**represents the value of some physical characteristic as a function of time, its derivative**

*h(t)***represents an instantaneous speed of change of this physical characteristic.**

*h'(t)*Using this logic, we can find an instantaneous speed of change of an

instantaneous speed of change, which would be a derivative of the

derivative, that is a second derivative of a base function

**.**

*h(t)*This is the logic behind analyzing how an instantaneous speed of change of an object's position itself changes with time.

*Acceleration*is an instantaneous speed of change of an

instantaneous speed of change of an object's position. By definition, it

is a vector with components equal to derivatives of corresponding

components of an instantaneous speed of change of an object's position.

Considering the latter is a vector with components equal to derivatives

of original coordinate functions,

*acceleration*is a vector

**represented by three second derivatives of the coordinate functions describing the object's position:**

*a***= {**

*a***}**

*x"(t), y"(t), z"(t)*A short note about terminology.

We distinguished

*velocity*(a vector) from

*speed*(its magnitude). In case of

*acceleration*there is no separate word to distinguish a

*vector of acceleration*from its magnitude. Both are called

*acceleration*, and it's the context that helps to distinguish one from another.

Usually, if the motion occurs in one dimension along a straight line in

one direction, there is no big difference between a vector and its

magnitude, and there should be no confusion. In two- and

three-dimensional cases with movement along some curve we will usually

consider

*acceleration*as a

*vector*.

Let's consider a few examples.

*Example 1*. An object is at rest at the origin of coordinates.

So,

**for all time moments**

*x(t)=y(t)=z(t)=0***.**

*t*Velocity

**of this movement is**

*v***for all time moments, since a derivative from a constant equals to**

*null-vector***.**

*0*Indeed,

**and**

*x'(t)=y'(t)=z'(t)=0***{**

*v =***} =**

*x'(t), y'(t), z'(t)*= {

**}.**

*0, 0, 0*Hence, its instantaneous speed

**(magnitude of the vector of velocity) is also**

*s***.**

*0*Acceleration, similarly, is a

**for all time moments as well, as**

*null-vector***{**

*a =***} =**

*x"(t), y"(t), z"(t)*= {

**}.**

*0, 0, 0*Its magnitude is, obviously,

**.**

*0**Example 2*. An object is at rest at a point with coordinates

**}.**

*3, 10, −6*So,

**for all time moments**

*x(t)=3; y(t)=10; z(t)=−6***.**

*t*Velocity

**of this movement is**

*v***for all time moments, since a derivative from a constant equals to**

*null-vector***.**

*0*Indeed,

**and**

*x'(t)=y'(t)=z'(t)=0***{**

*v =***} =**

*x'(t), y'(t), z'(t)*= {

**}.**

*0, 0, 0*Hence, its instantaneous speed

**(magnitude of the vector of velocity) is also**

*s***.**

*0*Acceleration, similarly, is a

**for all time moments as well, as**

*null-vector***{**

*a =***} =**

*x"(t), y"(t), z"(t)*= {

**}.**

*0, 0, 0*Its magnitude is, obviously,

**.**

*0**Example 3*. An object is in the uniform motion. At time

*t=0*it is at a point with coordinates

**}.**

*3, 10, −6*It moves along a straight line according to the following coordinate functions:

*x(t)=3+6·t;*

y(t)=10−8·t;

z(t)=−6+10·ty(t)=10−8·t;

z(t)=−6+10·t

for all time moments

**.**

*t*Velocity

**is vector with the following components:**

*v***.**

*x'(t)=6; y'(t)=−8; z'(t)=10*Therefore,

**{**

*v =***} =**

*x'(t), y'(t), z'(t)*= {

**}.**

*6, −8, 10*Hence, its instantaneous speed

**(magnitude of the vector of velocity) is**

*s***.**

*s = √(6)²+(−8)²+(10)² = 10√2*Acceleration is a

**for all time moments of time because the velocity is constant:**

*null-vector***{**

*a =***} =**

*x"(t), y"(t), z"(t)*= {

**}.**

*0, 0, 0*The magnitude of acceleration is, obviously,

**.**

*0**Example 4*. An object falls down from the Tower of Pisa, pulled by gravity, accelerating all the time. At time

*t=0*it is at the top of the tower at a point with coordinates

**}**

*0, 0, H***- some positive number that defines initial height of an object - the height of the Tower of Pisa.**

*H*It moves along a straight line down the Z-axis towards point

**}**

*0, 0, 0*

*x(t)=0;*

y(t)=0;

z(t)=H−g·t²/2y(t)=0;

z(t)=H−g·t²/2

(where

**is a positive constant related to strength of the gravity)**

*g*for all time moments

**.**

*t*Velocity

**is vector with the following components:**

*v***.**

*x'(t)=0; y'(t)=0; z'(t)=−g·t*Therefore,

**{**

*v =***} =**

*x'(t), y'(t), z'(t)*= {

**}.**

*0, 0, −g·t*Hence, its instantaneous speed

**(magnitude of the vector of velocity) is**

*s***.**

*s = √(0)²+(0)²+(−g·t)² = g·t*Acceleration

**is a vector with the following components:**

*a***{**

*a =***} =**

*x"(t), y"(t), z"(t)*= {

**}.**

*0, 0, −g*The magnitude of acceleration is, obviously,

**.**

*g*We can also calculate the time, when this object reaches the ground (that is, when

**). For this we just have to resolve the equation**

*z(t)=0***.**

*z(t) = H−g·t²/2 = 0*The solution is, obviously,

**.**

*t = √2H/g*
## No comments:

Post a Comment