User:
Translation
Description
The basis of mechanics lies in describing the position
Translation refers to the movement of an object over time, resulting in a change in its position
By establishing a function for an object\'s position over time, it becomes possible to predict its future motion and trajectory over time.
ID:(466, 0)
Average speed
Description
To estimate how an object moves, we need to know the path over time it follows. Hence, the ratio of the distance traveled and the time elapsed, which is defined as the average velocity, is introduced.
To measure this, we can use a system like the one shown in the image:
To determine the average velocity, two sensors must be placed to record the passage of an object at a distance $\Delta s$. The time difference in which the object passes in front of each sensor is then recorded as $\Delta t$. The average velocity is determined by dividing the distance traveled by the time elapsed.
The equation that describes the mean Speed ($\bar{v}$) with the distance traveled in a time ($\Delta s$) and the time elapsed ($\Delta t$) is the following:
| $ \bar{v} \equiv\displaystyle\frac{ \Delta s }{ \Delta t }$ |
It should be noted that the average velocity is an estimate of the actual velocity. The main problem lies in the fact that:
If the velocity changes during the elapsed time, the value of the average velocity can be very different from the average speed.
Additionally, there is a problem with the way the distance traveled is measured, as we work with two positions. This can lead to the situation where:
Since the path traveled is calculated from the difference between two positions, in cases where the movement is reversed during the elapsed time, the initial and final positions can be very similar. This can lead to an average velocity that is approximately zero, despite having traveled a 'long' distance.
Therefore, the key is to:
Determine the velocity in a sufficiently short elapsed time so that its variation is minimal.
ID:(470, 0)
Instant speed
Description
The speed ($v$) is defined as the displacement per unit of time. However, this concept is reduced to ERROR:5268.1 that exists during the interval of time considered.
The limitation of the average velocity is reflected in the assumption that an object instantly goes from rest to a given velocity. It is as if a bus that has just left the terminal suddenly reaches cruising speed, which is totally absurd. The velocity evolves, increases, decreases (due to a traffic light or picking up passengers), and slowly increases until it reaches a more or less constant value when traveling on the road. Therefore, a bus that normally travels on the road at around 100 km/h will take more than 8 hours to cover 800 km because the fluctuations in velocity must be taken into account. In the end, it will have taken 10 hours to cover 800 km traveling at an average speed of 80 km/h.
If we want to know the velocity at each moment, we must take a time interval so small that during this time, the velocity can be considered approximately constant. Thus, the average velocity estimated in this way is equivalent to the velocity that exists at the considered moment.
That's why we talk about instantaneous velocity.
ID:(16, 0)
Time path graph for constant speed with no initial time
Description
For the case of constant velocity without initial time, the position can be calculated using ,
| $ s_v = s_0 + v_0 t $ |
which corresponds to a line with:
• slope equal to the velocity $v_0$
• initial position $x_0$
as shown below:
ID:(2244, 0)
Time path graph for constant speed and initial time
Description
For the case of constant velocity and initial time, the position can be calculated using the values the position ($s$), the starting position ($s_0$), the constant velocity ($v_0$), the time ($t$), and the start Time ($t_0$) with the following equation:
| $ s = s_0 + v_0 ( t - t_0 )$ |
which corresponds to a straight line with:
• a slope equal to the constant velocity ($v_0$)
• a y-intercept at the starting position ($s_0$) for the start Time ($t_0$)
as illustrated below:
ID:(2243, 0)
Paradox of the body at rest
Description
When a body is said to be "at rest", it means that it is at rest relative to our reference frame or coordinate system. However, this "rest" is entirely relative, meaning that from a body that moves relative to our system, the body at "rest" is also in motion.
In this sense, there is no such thing as an "absolute rest", it only exists as something relative to a particular reference frame. Therefore, in general, all velocity measurements are relative to a particular reference frame.
For example, if a body appears to move very slowly, it only means that its velocity is very similar to the velocity of the reference frame in which the slow movement is observed.
ID:(4405, 0)
Speed as a derivative
Description
If we take ERROR:5264.1 with ERROR:9899.1 ($s(t)$) and observe a point at a future time $t+\Delta t$ with a position $s(t+\Delta t)$, we can estimate the velocity as the distance traveled in the time $\Delta t$:
$s(t+\Delta t)-s(t)$
the speed ($v$) can be calculated by dividing the distance traveled by the elapsed time:
$v\sim\displaystyle\frac{s(t+\Delta t)-s(t)}{\Delta t}$
As the value of $\Delta t$ becomes smaller, the calculated velocity approaches the tangent to the position curve at the time:
This generalizes what had already been seen for the case of constant velocity.
ID:(1638, 0)
Velocity as slope of position curve
Description
If displacement is graphed as a line between the origin O and point A:
it can be seen that a path has been traveled over a period of time. Therefore, the slope of the graph of path vs elapsed time corresponds to velocity.
If the slope is steeper, it means that a path is covered in less time, which corresponds to a higher velocity.
If the slope is flatter, it means that a path is covered in more time, which corresponds to a lower velocity.
ID:(2239, 0)
Negative slope
Description
When a segment on a graph has a negative slope, as shown below:
it represents a situation in which the object returned from position B to C, which is a distance of zero from the origin. In other words, negative slopes correspond to traveling in the opposite direction, not moving away but getting closer to the origin.
ID:(2245, 0)
Path as integral of speed
Description
The integral of the speed ($v$) corresponds to the area under the curve that defines this function. Therefore, the integral of the velocity between the start Time ($t_0$) and the time ($t$) corresponds to the distance traveled between the starting position ($s_0$) and the position ($s$).
Therefore, we have:
| $ s = s_0 +\displaystyle\int_{t_0}^t v d\tau$ |
Which is represented in the following graph:
I walk as an area under the speed and time curve.
ID:(2242, 0)
Time path diagram with horizontal segment
Description
A second type of case is horizontal segments in the distance vs time graph:
If we observe segment AB, we will notice that despite the elapsed time, the distance has not changed. This means that the object is at rest. Therefore, horizontal segments, which correspond to zero slope, correspond to periods when the velocity is zero.
ID:(2241, 0)
Velocity in more than one dimension
Description
The velocity of an object in a system with more than one dimension is traditionally defined using a vector. This vector originates from the object and is oriented in the direction in which the object is moving.
ID:(2240, 0)
Concepts
Description
Variables
Calculations
to 
,
then, select the variable: 
to 
Calculations
Variable
Given
Calculate
Target :
Equation
To be used
Equations
If we consider the distance traveled as the difference in position between time $t+\Delta t$ and time $t$:
$\Delta s = s(t+\Delta t)-s(t)$
and take $\Delta t$ as the elapsed time, then in the limit of infinitesimally short times, the average velocity can be expressed as:
$v_m=\displaystyle\frac{\Delta s}{\Delta t}=\displaystyle\frac{s(t+\Delta t)-s(t)}{\Delta t}\rightarrow \lim_{\Delta t\rightarrow 0}\displaystyle\frac{s(t+\Delta t)-s(t)}{\Delta t}=\displaystyle\frac{ds}{dt}$
This last expression corresponds to the derivative of the position function $s(t)$:
| $ v =\displaystyle\frac{ d s }{ d t }$ |
which in turn is the slope of the graphical representation of this function over time.
(ID 3153)
With the distance traveled in a time ($\Delta s$) it is with the position ($s$) and the starting position ($s_0$):
| $ \Delta s = s - s_0 $ |
and the time elapsed ($\Delta t$) is with the time ($t$) and the start Time ($t_0$):
| $ \Delta t \equiv t - t_0 $ |
The equation for average velocity:
| $ v_0 \equiv\displaystyle\frac{ \Delta s }{ \Delta t }$ |
can be written as:
$v_0 = \bar{v} = \displaystyle\frac{\Delta s}{\Delta t} = \displaystyle\frac{s - s_0}{t - t_0}$
thus, solving for it we get:
| $ s = s_0 + v_0 ( t - t_0 )$ |
(ID 3154)
As a vector, velocity can be expressed as an array of its different components:
$\vec{v}=\begin{pmatrix}v_x\v_y\v_z\end{pmatrix}$
And its derivative can be expressed as the derivative of each of its components:
$\displaystyle\frac{d\vec{v}}{dt}=\begin{pmatrix}\displaystyle\frac{d v_x}{dt}\displaystyle\frac{d v_y}{dt}\displaystyle\frac{d v_z}{dt}\end{pmatrix}=\begin{pmatrix}a_x\a_y\a_z\end{pmatrix}=\vec{a}$
Thus, in general, the instantaneous velocity in more than one dimension is a vector with components in each of the directions:
| $ \vec{ v } =\displaystyle\frac{d \vec{s} }{d t }$ |
(ID 4354)
Examples
The basis of mechanics lies in describing the position
Translation refers to the movement of an object over time, resulting in a change in its position
By establishing a function for an object\'s position over time, it becomes possible to predict its future motion and trajectory over time.
(ID 466)
To estimate how an object moves, we need to know the path over time it follows. Hence, the ratio of the distance traveled and the time elapsed, which is defined as the average velocity, is introduced.
To measure this, we can use a system like the one shown in the image:
To determine the average velocity, two sensors must be placed to record the passage of an object at a distance $\Delta s$. The time difference in which the object passes in front of each sensor is then recorded as $\Delta t$. The average velocity is determined by dividing the distance traveled by the time elapsed.
The equation that describes the mean Speed ($\bar{v}$) with the distance traveled in a time ($\Delta s$) and the time elapsed ($\Delta t$) is the following:
| $ \bar{v} \equiv\displaystyle\frac{ \Delta s }{ \Delta t }$ |
It should be noted that the average velocity is an estimate of the actual velocity. The main problem lies in the fact that:
If the velocity changes during the elapsed time, the value of the average velocity can be very different from the average speed.
Additionally, there is a problem with the way the distance traveled is measured, as we work with two positions. This can lead to the situation where:
Since the path traveled is calculated from the difference between two positions, in cases where the movement is reversed during the elapsed time, the initial and final positions can be very similar. This can lead to an average velocity that is approximately zero, despite having traveled a 'long' distance.
Therefore, the key is to:
Determine the velocity in a sufficiently short elapsed time so that its variation is minimal.
(ID 470)
The speed ($v$) is defined as the displacement per unit of time. However, this concept is reduced to ERROR:5268.1 that exists during the interval of time considered.
The limitation of the average velocity is reflected in the assumption that an object instantly goes from rest to a given velocity. It is as if a bus that has just left the terminal suddenly reaches cruising speed, which is totally absurd. The velocity evolves, increases, decreases (due to a traffic light or picking up passengers), and slowly increases until it reaches a more or less constant value when traveling on the road. Therefore, a bus that normally travels on the road at around 100 km/h will take more than 8 hours to cover 800 km because the fluctuations in velocity must be taken into account. In the end, it will have taken 10 hours to cover 800 km traveling at an average speed of 80 km/h.
If we want to know the velocity at each moment, we must take a time interval so small that during this time, the velocity can be considered approximately constant. Thus, the average velocity estimated in this way is equivalent to the velocity that exists at the considered moment.
That's why we talk about instantaneous velocity.
(ID 16)
The mean Speed ($\bar{v}$) can be calculated from the distance traveled in a time ($\Delta s$) and the time elapsed ($\Delta t$) using:
| $ \bar{v} \equiv\displaystyle\frac{ \Delta s }{ \Delta t }$ |
(ID 3152)
In general, velocity should be understood as a three-dimensional entity, that is, a vector. Its position is described by a position vector
| $ v =\displaystyle\frac{ d s }{ d t }$ |
This allows for the generalization of velocity:
| $ \vec{ v } =\displaystyle\frac{d \vec{s} }{d t }$ |
(ID 4354)
The instantaneous ERROR:6029,0, determined by the relationship between the infinitesimal distance traveled ($ds$) and the infinitesimal Variation of Time ($dt$), provides a more accurate estimate of the actual velocity at any moment of the time ($t$), compared to the mean Speed ($\bar{v}$), which is calculated from the distance traveled in a time ($\Delta s$) and the time elapsed ($\Delta t$) using the equation:
| $ \bar{v} \equiv\displaystyle\frac{ \Delta s }{ \Delta t }$ |
This is achieved through the derivative of position with respect to time, i.e.,:
| $ v =\displaystyle\frac{ d s }{ d t }$ |
Thus, the instantaneous velocity the speed ($v$) of the position ($s$) is known at any instant of the time ($t$) with greater precision.
(ID 3153)
(ID 2244)
(ID 2243)
(ID 4405)
If the speed is constant, the velocity will be equal to the initial Speed ($v_0$). In this case, the distance traveled as a function of time can be calculated using the difference between the position ($s$) and the starting position ($s_0$), divided by the difference between the time ($t$) and the start Time ($t_0$):
| $ s = s_0 + v_0 ( t - t_0 )$ |
The corresponding equation defines a straight line in space-time.
(ID 3154)
If we take ERROR:5264.1 with ERROR:9899.1 ($s(t)$) and observe a point at a future time $t+\Delta t$ with a position $s(t+\Delta t)$, we can estimate the velocity as the distance traveled in the time $\Delta t$:
$s(t+\Delta t)-s(t)$
the speed ($v$) can be calculated by dividing the distance traveled by the elapsed time:
$v\sim\displaystyle\frac{s(t+\Delta t)-s(t)}{\Delta t}$
As the value of $\Delta t$ becomes smaller, the calculated velocity approaches the tangent to the position curve at the time:
This generalizes what had already been seen for the case of constant velocity.
(ID 1638)
(ID 2239)
(ID 2245)
The integral of the speed ($v$) corresponds to the area under the curve that defines this function. Therefore, the integral of the velocity between the start Time ($t_0$) and the time ($t$) corresponds to the distance traveled between the starting position ($s_0$) and the position ($s$).
Therefore, we have:
| $ s = s_0 +\displaystyle\int_{t_0}^t v d\tau$ |
Which is represented in the following graph:
I walk as an area under the speed and time curve.
(ID 2242)
(ID 2241)
The velocity of an object in a system with more than one dimension is traditionally defined using a vector. This vector originates from the object and is oriented in the direction in which the object is moving.
(ID 2240)
ID:(35, 0)