Unizor - Physics4Teens - Mechanics - Kinematics - Acceleration

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



Acceleration



Recall that, if x(t), y(t) and z(t) are coordinate functions of the moving object, the derivatives of these functions x'(t), y'(t) and z'(t) represent components of a velocity of this moving object, that is a vector of instantaneous change of an object's position.



Let's generalize. If function h(t) 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.

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

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, x(t)=y(t)=z(t)=0 for all time moments t.

Velocity v of this movement is null-vector for all time moments, since a derivative from a constant equals to 0.

Indeed, x'(t)=y'(t)=z'(t)=0 and v = {x'(t), y'(t), z'(t)} =

= {0, 0, 0}.

Hence, its instantaneous speed s (magnitude of the vector of velocity) is also 0.

Acceleration, similarly, is a null-vector for all time moments as well, as 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, x(t)=3; y(t)=10; z(t)=−6 for all time moments t.

Velocity v of this movement is null-vector for all time moments, since a derivative from a constant equals to 0.

Indeed, x'(t)=y'(t)=z'(t)=0 and v = {x'(t), y'(t), z'(t)} =

= {0, 0, 0}.

Hence, its instantaneous speed s (magnitude of the vector of velocity) is also 0.

Acceleration, similarly, is a null-vector for all time moments as well, as 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·t

for all time moments t.

Velocity v is vector with the following components:

x'(t)=6; y'(t)=−8; z'(t)=10.

Therefore, v = {x'(t), y'(t), z'(t)} =

= {6, −8, 10}.

Hence, its instantaneous speed s (magnitude of the vector of velocity) is

s = √(6)²+(−8)²+(10)² = 10√2.

Acceleration is a null-vector for all time moments of time because the velocity is constant:

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

It moves along a straight line down the Z-axis towards point {0, 0, 0} on the ground, according to the following coordinate functions:

x(t)=0;

y(t)=0;

z(t)=H−g·t²/2

(where g is a positive constant related to strength of the gravity)
for all time moments t.

Velocity v is vector with the following components:

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 s (magnitude of the vector of velocity) is

s = √(0)²+(0)²+(−g·t)² = g·t.

Acceleration a is a vector with the following components:

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 z(t)=0). For this we just have to resolve the equation z(t) = H−g·t²/2 = 0.

The solution is, obviously, t = √2H/g .