dynamics of rigid bodies

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Yanyan Badeo
Yanyan Badeo's picture
dynamics of rigid bodies

will you please help me with these?

when the acceleration is zero (a=0)
a) s=vt

when the acceleration is constant (uniformly accelerated motion)

a) v = Vot + at
b) s = Vot + 1/2 at
c) v^2 = Vo ^2 + 2as

acceleration (a)

a) a= d^2x/dt = d^2s/ dt

b) a= vdv /dx = vdv / ds

fitzmerl duron
fitzmerl duron's picture

To derive those formulas you've presented...here's how....

$\mathit{\mathbf{For \space the \space first \space part}}$


To derive $s = vt$, you can do a dimensional analysis. Let's say $s$ is a distance expressed in meters, $v$ is speed expressed in meters per second and $t$ is time measured in seconds. Then doing this:

$$s = vt$$ $$meters = \left(\frac{meters}{second}\right)(seconds)$$ $$meters = meters$$

Actually, you can't derive the equation $s = vt$ because it is a basic formula. It is composed of fundamental units of length and time, but it can be used to create richer equations/derived equations such as those equations below.

$\mathit{\mathbf{For \space the \space second \space part}}$


To derive the equation $v=v_o+at$, we start by using the average acceleration. The average acceleration $\bar a$ during the time interval $\Delta t$ is the change in velocity $\Delta v$ divided by $\Delta t$. In symbols:

$$\bar a = \frac{\Delta v}{\Delta t} = \frac{v_f-v_i}{t_f-t_i}------> (1)$$

where $v_f$ is final velocity, $v_i$ is initial velocity, $t_f$ is final time and $t_i$ is the initial time.

The bar sign above $a$ signifies that the average acceleration above is an arithmetic average of that acceleration.

Most objects move with constant acceleration, such as objects in free fall near Earth's surface. When objects move with constant acceleration, the instantaneous acceleration at any point in a time interval is equal to the value of the average acceleration over the entire time interval. Imagine this instantaneous acceleration as freezing the falling ball at arbitrary time and discerning the acceleration of that ball at that moment ALA bullet time in the film The Matrix....hehehe

Because if acceleration is constant, instantaneous acceleration (acceleration measured in a very short time) would have the same vaue as average acceleration (acceleration measured in longer times).

Rewriting the equation $(1)$ to get the instantaneous acceleration, and we know that average acceleration $=$ instantaneous acceleration if accelearation is constant, we get...

$$ a = \frac{\Delta v}{\Delta t} = \frac{v_f-v_i}{t_f-t_i}------------> (2)$$

We didn't write the bar sign atop $a$ because it is now an instantaneous acceleration. The value of instantaneous acceleration is not an arithmetic average of acceleration because this acceleration is measured in very quite brief time.

We now let $v_f = v$ ($v$ is velocity at any time $t$), $v_i = v_o$ ($v_o$ is initial velocity at $t = 0$), $t_f$ be any time $t$ and $t_i$ be zero. Substituting these values to $(2)$, we get...

$$ a = \frac{v_f-v_i}{t_f-t_i}$$ $$ a = \frac{v-v_o}{t-0}$$ $$ a = \frac{v-v_o}{t}$$

We now get the expression of velocity at any time $v$ as...
$$\color{green}{v = v_o +at}-------->(3)$$


To derive $s = v_ot + \frac{1}{2}at^2$, we start the notion that velocity is increasing or decreasing uniformly with time because of constant acceleration. We can express the average velocity $\bar v$ in any time interval as the arithmetic average of the initial velocity $v_o$ and the final velocity $v$. In symbols:

$$\bar v = \frac{v_o + v}{2}---------->(4)$$

The equation $(4)$ is valid if acceleration is constant.

We can now use $(4)$ to get $s = v_ot + \frac{1}{2}at^2$. Using the basic formula $\Delta x = x_f-x_i = s = (v_f-v_i)(t_f-t_i)$, where $s$ is displacement, and letting $x_f = x$ ($x_f$ is the final point), $x_i = x_o$ ($x_i$ is the initial/starting point), $v_f = v$, $v_i = 0$, $t_i = 0$, $t_f = t$, we now get...

$$\Delta x = \bar vt = \left( \frac{v_o+v}{2}\right)t$$ $$\Delta x = \frac{1}{2}(v_o + v)t------->(5)$$ $$\Delta x = \frac{1}{2}(v_o + (v_o + at))t$$ $$\color{green}{\Delta x = s = v_ot + \frac{1}{2}at^2}------>(6)$$

The equation $\color{green}{s = v_ot + \frac{1}{2}at^2}$ gives the displacement as a function of time.


To derive the expression $v^2 = v_o^2 +2as$, we need to get the time $t$ in equation $(3)$ and substituting into equation $(5)$, we get...

$$s = \frac{1}{2}(v_o + v)t--------->(5)$$ $$s = \frac{1}{2}(v_o + v)\left( \frac{v-v_o}{a}\right) = \frac{v^2 - v_o^2}{2a}$$ $$\color{green}{v^2 = v_o^2 + 2as} -------->(7)$$

The expression $v^2 = v_o^2 + 2as$ gives the velocity as a function of displacement.

$\mathit{\mathbf{For \space the \space third \space part}}$

To derive the acceleration $a$ as $\frac{d^2x}{dt^2}$ or $\frac{d^2s}{dt^2}$, here's how...

By definition, the acceleration of a body $M$ is defined as the first derivative of the velocity of $M$ relative to a given point of reference with respect to time:


Also by definition, the velocity of M is defined as the first derivative of the displacement of M from a given point of reference with respect to time:


That is:


Hence the result by definition of the second derivative $\frac{d^2s}{dt^2}$.

Additional info:

The speed $v$ is the first derivative of displacement. In symbols: $v = \frac{dx}{dt}$

The acceleration $a$ is the second derivative of displacement. In symbols: $a = \frac{d^2x}{dt^2}$

The jerk $\vec j$ is the third derivative of displacement. In symbols: $\vec j = \frac{d^3x}{dt^3}$

The jounce $j$ is the fourth derivative of displacement. In symbols: $j = \frac{d^3x}{dt^3}$

Nakatama tayo ng dalawang ibon gamit isang bato sa ikatlong parte...hihihhi....

Alternate solutions are highly encouraged....

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