Monday, April 23, 2018

Unizor - Physics - Mechanics - Kinematics - Constant Acceleration

Notes to a video lecture on

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} is characterized by the following equations for position functions x(t), y(t) and z(t):

x"(t) = a

y"(t) = b

z"(t) = c

where a, b and c are constants
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 P(t) are coordinate functions x(t), y(t) and 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 A is:

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 + vx(0)

y'(t) = b·t + vy(0)

z'(t) = c·t + vz(0)

where vx(0), vy(0) and vz(0) are some constants.

We have chosen the names of these constants to emphasize their meaning - components of velocity vector V(t) at time t=0:

x'(0) = vx(0)

y'(0) = vy(0)

z'(0) = vz(0)

or, in vector form,

V(0) = {vx(0), vy(0). vz(0)}

Continuing the vector representation of our equations of motion, we can express velocity vector V(t) as a linear function of time using acceleration vector A and initial velocity vector V(0) as parameters:

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 + vx(0)·t + x0

y(t) = b·t²/2 + vy(0)·t + y0

z(t) = c·t²/2 + vz(0)·t + z0

where x0, y0 and z0 are some constants equal to position of our object at time t=0:

x(0) = x0

y(0) = y0

z(0) = z0

In vector form the above equations of motion can be represented shorter:

P(t) = A·t²/2 + V0·t + P0,

where V0 = V(0) - initial velocity vector and P0 = P(0) - initial position.

Obviously, the second derivatives of these three quadratic functions above correspondingly equal to a, b and c - components of our constant acceleration vector of motion.

How to determine constants vx(0), vy(0), vz(0), x0, y0 and z0 that participate in the equations for position of our object?

Well, we cannot without additional information.

As we have determined, V0 = {vx(0), vy(0), vz(0)} is velocity vector at the beginning of motion t=0 and P0 = {x0, y0, z0} is a point, where the motion starts. Knowledge of this 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), y(t) and 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

a = vx(0)

b = vy(0)

c = vz(0)

Or in vector form

A = V0

In such a case

x(t) = a·(t²/2 + k·t) + x0

y(t) = b·(t²/2 + k·t) + y0

z(t) = c·(t²/2 + k·t) + z0

Or in vector form

P(t) = A·(t²/2 + k·t) + P0

The equations above define a trajectory collinear to vector A = {a, b, c}, and the proof is analogous to that we offered in the previous lecture about 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 y(t)=z(t)=0 and y'(t)=z'(t)=0 for all moments of time t. In particular, velocity vector looks now as {x'(t), 0, 0} and acceleration vector looks like {a, 0, 0}.

For these cases traditionally the letter v is used for the X-component of the velocity vector. So, we will use in these cases the symbol v(t) for X-component of the velocity vector x'(t).

The only non-trivial equation of motion then will be about x(t):

x(t) = a·t²/2 + v0·t + x0

(where x0=x(0) is initial position, v0=v(0) is initial speed of an object at the start of motion and a - its constant acceleration).

If the origin of the system of coordinates coincides with the initial position of the object, x0=0 and the equation of motion looks simpler:

x(t) = a·t²/2 + v0·t

If, in addition, the object is at rest in the beginning of motion, v0=0 and the equation of motion looks really simple:

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 v(t) instead of velocity.


An object moving from the initial position P0 ={x0, y0, z0} with initial velocity vector V0 ={vx(0), vy(0), vz(0)} and constant acceleration vector A ={a, b, c} moves along a trajectory with the following equations of motion:

x(t) = a·t²/2 + vx(0)·t + x0

y(t) = b·t²/2 + vy(0)·t + y0

z(t) = c·t²/2 + vz(0)·t + z0

Or, in vector form,

P(t) = A·t²/2 + V0·t + P0

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 + v0·t + x0

y(t) = 0

z(t) = 0

If, in addition, the origin of coordinates coincides with the point of beginning of motion, that is x0=0, the equations are simplified even further:

x(t) = a·t²/2 + v0·t

y(t) = 0

z(t) = 0

Finally, if, in addition, the starting velocity is zero, that is v0=0, the equations are simplified even further:

x(t) = a·t²/2

y(t) = 0

z(t) = 0

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