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Intercept at constant acceleration
Storyboard 
Objects can intersect when they coincide in position at the same moment. To achieve this, they must move from their respective starting points and velocities with accelerations that allow them to coincide in position and time at the end of the journey.
ID:(1412, 0)
Variation in speed and duration
Concept
In a scenario of motion involving two bodies, the first one alters its velocity by the speed difference of the first body ($\Delta v_1$) during ($$) with the first body acceleration ($a_1$).
| $ a_1 \equiv\displaystyle\frac{ \Delta v_1 }{ \Delta t_1 }$ |
Subsequently, the second body advances, altering its velocity by the second body speed difference ($\Delta v_2$) during a time span of the travel time of second object ($\Delta t_2$) with the second body acceleration ($a_2$).
| $ a_2 \equiv\displaystyle\frac{ \Delta v_2 }{ \Delta t_2 }$ |
When represented graphically, we obtain a velocity-time diagram as shown below:
The key here is that the values the speed difference of the first body ($\Delta v_1$) and the second body speed difference ($\Delta v_2$), and the values the travel time of first object ($\Delta t_1$) and the travel time of second object ($\Delta t_2$), are such that both bodies coincide in place and time.
ID:(12512, 0)
Speed and intersection times
Concept
In the case of two bodies, the motion of the first one can be described by a function involving the points the initial time of first object ($t_1$), the intersection time ($t$), the initial velocity of the first body ($v_{01}$), and the final velocity of the first body ($v_1$), represented by a line with a slope of the first body acceleration ($a_1$):
| $ v_1 = v_{01} + a_1 ( t - t_1 )$ |
For the motion of the second body, defined by the points the initial velocity of the second body ($v_{02}$), the final velocity of the second body ($v_2$), the initial time of second object ($t_2$), and the intersection time ($t$), a second line with a slope of the second body acceleration ($a_2$) is employed:
| $ v_2 = v_{02} + a_2 ( t - t_2 )$ |
This is represented as:
ID:(12515, 0)
Evolution of the position of the bodies
Concept
In the case of a two-body motion, the position where the trajectory of the first body ends coincides with that of the second body at the intersection position ($s$).
Similarly, the time at which the trajectory of the first body ends coincides with that of the second body at the intersection time ($t$).
For the first body, the intersection position ($s$) depends on the initial position of first object ($s_1$), the initial velocity of the first body ($v_{01}$), the first body acceleration ($a_1$), the initial time of first object ($t_1$), as follows:
| $ s = s_1 + v_{01} ( t - t_1 )+\displaystyle\frac{1}{2} a_1 ( t - t_1 )^2$ |
While for the second body, the intersection position ($s$) depends on the initial position of second object ($s_2$), the initial velocity of the second body ($v_{02}$), the second body acceleration ($a_2$), the initial time of second object ($t_2$), as follows:
| $ s = s_2 + v_{02} ( t - t_2 )+\displaystyle\frac{1}{2} a_2 ( t - t_2 )^2$ |
This is represented as:
ID:(12513, 0)
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Equations
$ a_1 \equiv\displaystyle\frac{ \Delta v_1 }{ \Delta t_1 }$
a_m = Dv / Dt
$ a_2 \equiv\displaystyle\frac{ \Delta v_2 }{ \Delta t_2 }$
a_m = Dv / Dt
$ \Delta s_1 \equiv s - s_1 $
Ds = s - s_0
$ \Delta s_2 \equiv s - s_2 $
Ds = s - s_0
$ \Delta t_1 \equiv t - t_1 $
Dt = t - t_0
$ \Delta t_2 \equiv t - t_2 $
Dt = t - t_0
$ \Delta v_1 \equiv v_1 - v_{01} $
Dv = v - v_0
$ \Delta v_2 \equiv v_2 - v_{02} $
Dv = v - v_0
$ s = s_1 + v_{01} ( t - t_1 )+\displaystyle\frac{1}{2} a_1 ( t - t_1 )^2$
s = s_0 + v_0 * ( t - t_0 )+ a_0 *( t - t_0 )^2/2
$ s = s_2 + v_{02} ( t - t_2 )+\displaystyle\frac{1}{2} a_2 ( t - t_2 )^2$
s = s_0 + v_0 * ( t - t_0 )+ a_0 *( t - t_0 )^2/2
$ s = s_1 +\displaystyle\frac{ v_1 ^2- v_{01} ^2}{2 a_1 }$
s = s_0 +( v ^2- v_0 ^2)/(2* a_0 )
$ s = s_2 +\displaystyle\frac{ v_2 ^2- v_{02} ^2}{2 a_2 }$
s = s_0 +( v ^2- v_0 ^2)/(2* a_0 )
$ v_1 = v_{01} + a_1 ( t - t_1 )$
v = v_0 + a_0 *( t - t_0 )
$ v_2 = v_{02} + a_2 ( t - t_2 )$
v = v_0 + a_0 *( t - t_0 )
ID:(15402, 0)
Speed variation (1)
Equation
Acceleration corresponds to the change in velocity per unit of time.
Therefore, it is necessary to define the speed Diference ($\Delta v$) in terms of the speed ($v$) and the initial Speed ($v_0$) as follows:
| $ \Delta v_1 \equiv v_1 - v_{01} $ |
| $ \Delta v \equiv v - v_0 $ |
ID:(4355, 1)
Speed variation (2)
Equation
Acceleration corresponds to the change in velocity per unit of time.
Therefore, it is necessary to define the speed Diference ($\Delta v$) in terms of the speed ($v$) and the initial Speed ($v_0$) as follows:
| $ \Delta v_2 \equiv v_2 - v_{02} $ |
| $ \Delta v \equiv v - v_0 $ |
ID:(4355, 2)
Elapsed time (1)
Equation
To describe the motion of an object, we need to calculate the time elapsed ($\Delta t$). This magnitude is obtained by measuring the start Time ($t_0$) and the the time ($t$) of said motion. The duration is determined by subtracting the initial time from the final time:
| $ \Delta t_1 \equiv t - t_1 $ |
| $ \Delta t \equiv t - t_0 $ |
ID:(4353, 1)
Elapsed time (2)
Equation
To describe the motion of an object, we need to calculate the time elapsed ($\Delta t$). This magnitude is obtained by measuring the start Time ($t_0$) and the the time ($t$) of said motion. The duration is determined by subtracting the initial time from the final time:
| $ \Delta t_2 \equiv t - t_2 $ |
| $ \Delta t \equiv t - t_0 $ |
ID:(4353, 2)
Mean acceleration (1)
Equation
The proportion in which the variation of velocity over time is defined as the mean Acceleration ($\bar{a}$). To measure it, it is necessary to observe the speed Diference ($\Delta v$) and the time elapsed ($\Delta t$).
One common method for measuring average acceleration involves using a stroboscopic lamp that illuminates the object at defined intervals. By taking a photograph, one can determine the distance traveled by the object in that time. By calculating two consecutive velocities, one can determine their variation and, with the time elapsed between the photos, the average acceleration.
The equation that describes average acceleration is as follows:
| $ a_1 \equiv\displaystyle\frac{ \Delta v_1 }{ \Delta t_1 }$ |
| $ \bar{a} \equiv\displaystyle\frac{ \Delta v }{ \Delta t }$ |
The definition of the mean Acceleration ($\bar{a}$) is considered as the relationship between the speed Diference ($\Delta v$) and the time elapsed ($\Delta t$). That is,
| $ \Delta v \equiv v - v_0 $ |
and
| $ \Delta t \equiv t - t_0 $ |
The relationship between both is defined as the centrifuge Acceleration ($a_c$)
| $ \bar{a} \equiv\displaystyle\frac{ \Delta v }{ \Delta t }$ |
within this time interval.
It is important to note that average acceleration is an estimation of actual acceleration.
The main problem is that if acceleration varies during the elapsed time, the value of the average acceleration may differ greatly from the mean acceleration
.
Therefore, the key is to
Determine acceleration over a sufficiently short period of time to minimize variation.
ID:(3678, 1)
Mean acceleration (2)
Equation
The proportion in which the variation of velocity over time is defined as the mean Acceleration ($\bar{a}$). To measure it, it is necessary to observe the speed Diference ($\Delta v$) and the time elapsed ($\Delta t$).
One common method for measuring average acceleration involves using a stroboscopic lamp that illuminates the object at defined intervals. By taking a photograph, one can determine the distance traveled by the object in that time. By calculating two consecutive velocities, one can determine their variation and, with the time elapsed between the photos, the average acceleration.
The equation that describes average acceleration is as follows:
| $ a_2 \equiv\displaystyle\frac{ \Delta v_2 }{ \Delta t_2 }$ |
| $ \bar{a} \equiv\displaystyle\frac{ \Delta v }{ \Delta t }$ |
The definition of the mean Acceleration ($\bar{a}$) is considered as the relationship between the speed Diference ($\Delta v$) and the time elapsed ($\Delta t$). That is,
| $ \Delta v \equiv v - v_0 $ |
and
| $ \Delta t \equiv t - t_0 $ |
The relationship between both is defined as the centrifuge Acceleration ($a_c$)
| $ \bar{a} \equiv\displaystyle\frac{ \Delta v }{ \Delta t }$ |
within this time interval.
It is important to note that average acceleration is an estimation of actual acceleration.
The main problem is that if acceleration varies during the elapsed time, the value of the average acceleration may differ greatly from the mean acceleration
.
Therefore, the key is to
Determine acceleration over a sufficiently short period of time to minimize variation.
ID:(3678, 2)
Speed by constant acceleration (1)
Equation
If the constant Acceleration ($a_0$), then the mean Acceleration ($\bar{a}$) is equal to the value of acceleration, that is,
| $ a_0 = \bar{a} $ |
.
In this case, the speed ($v$) as a function of the time ($t$) can be calculated by considering that it is associated with the difference between the speed ($v$) and the initial Speed ($v_0$), as well as the time ($t$) and the start Time ($t_0$).
| $ v_1 = v_{01} + a_1 ( t - t_1 )$ |
| $ v = v_0 + a_0 ( t - t_0 )$ |
In the case where the constant Acceleration ($a_0$) equals the mean Acceleration ($\bar{a}$), it will be equal to
| $ a_0 = \bar{a} $ |
.
Therefore, considering the speed Diference ($\Delta v$) as
| $ \Delta v \equiv v - v_0 $ |
and the time elapsed ($\Delta t$) as
| $ \Delta t \equiv t - t_0 $ |
,
the equation for the constant Acceleration ($a_0$)
| $ \bar{a} \equiv\displaystyle\frac{ \Delta v }{ \Delta t }$ |
can be written as
$a_0 = \bar{a} = \displaystyle\frac{\Delta v}{\Delta t} = \displaystyle\frac{v - v_0}{t - t_0}$
and by rearranging, we obtain
| $ v = v_0 + a_0 ( t - t_0 )$ |
.
This equation thus represents a straight line in velocity-time space.
ID:(3156, 1)
Speed by constant acceleration (2)
Equation
If the constant Acceleration ($a_0$), then the mean Acceleration ($\bar{a}$) is equal to the value of acceleration, that is,
| $ a_0 = \bar{a} $ |
.
In this case, the speed ($v$) as a function of the time ($t$) can be calculated by considering that it is associated with the difference between the speed ($v$) and the initial Speed ($v_0$), as well as the time ($t$) and the start Time ($t_0$).
| $ v_2 = v_{02} + a_2 ( t - t_2 )$ |
| $ v = v_0 + a_0 ( t - t_0 )$ |
In the case where the constant Acceleration ($a_0$) equals the mean Acceleration ($\bar{a}$), it will be equal to
| $ a_0 = \bar{a} $ |
.
Therefore, considering the speed Diference ($\Delta v$) as
| $ \Delta v \equiv v - v_0 $ |
and the time elapsed ($\Delta t$) as
| $ \Delta t \equiv t - t_0 $ |
,
the equation for the constant Acceleration ($a_0$)
| $ \bar{a} \equiv\displaystyle\frac{ \Delta v }{ \Delta t }$ |
can be written as
$a_0 = \bar{a} = \displaystyle\frac{\Delta v}{\Delta t} = \displaystyle\frac{v - v_0}{t - t_0}$
and by rearranging, we obtain
| $ v = v_0 + a_0 ( t - t_0 )$ |
.
This equation thus represents a straight line in velocity-time space.
ID:(3156, 2)
Path traveled with constant acceleration (1)
Equation
In the case of ($$), the speed ($v$) varies linearly with the time ($t$), using the initial Speed ($v_0$) and the start Time ($t_0$):
| $ v = v_0 + a_0 ( t - t_0 )$ |
Thus, the area under this line can be calculated, yielding the distance traveled in a time ($\Delta s$). Combining this with the starting position ($s_0$), we can calculate the position ($s$), resulting in:
| $ s = s_1 + v_{01} ( t - t_1 )+\displaystyle\frac{1}{2} a_1 ( t - t_1 )^2$ |
| $ s = s_0 + v_0 ( t - t_0 )+\displaystyle\frac{1}{2} a_0 ( t - t_0 )^2$ |
In the case of the constant Acceleration ($a_0$), the speed ($v$) as a function of the time ($t$) forms a straight line passing through the start Time ($t_0$) and the initial Speed ($v_0$), defined by the equation:
| $ v = v_0 + a_0 ( t - t_0 )$ |
Since the distance traveled in a time ($\Delta s$) represents the area under the velocity-time curve, we can sum the contributions of the rectangle:
$v_0(t-t_0)$
and the triangle:
$\displaystyle\frac{1}{2}a_0(t-t_0)^2$
To obtain the distance traveled in a time ($\Delta s$) with the position ($s$) and the starting position ($s_0$), resulting in:
| $ \Delta s \equiv s - s_0 $ |
Therefore:
| $ s = s_0 + v_0 ( t - t_0 )+\displaystyle\frac{1}{2} a_0 ( t - t_0 )^2$ |
This corresponds to the general form of a parabola.
ID:(3157, 1)
Path traveled with constant acceleration (2)
Equation
In the case of ($$), the speed ($v$) varies linearly with the time ($t$), using the initial Speed ($v_0$) and the start Time ($t_0$):
| $ v = v_0 + a_0 ( t - t_0 )$ |
Thus, the area under this line can be calculated, yielding the distance traveled in a time ($\Delta s$). Combining this with the starting position ($s_0$), we can calculate the position ($s$), resulting in:
| $ s = s_2 + v_{02} ( t - t_2 )+\displaystyle\frac{1}{2} a_2 ( t - t_2 )^2$ |
| $ s = s_0 + v_0 ( t - t_0 )+\displaystyle\frac{1}{2} a_0 ( t - t_0 )^2$ |
In the case of the constant Acceleration ($a_0$), the speed ($v$) as a function of the time ($t$) forms a straight line passing through the start Time ($t_0$) and the initial Speed ($v_0$), defined by the equation:
| $ v = v_0 + a_0 ( t - t_0 )$ |
Since the distance traveled in a time ($\Delta s$) represents the area under the velocity-time curve, we can sum the contributions of the rectangle:
$v_0(t-t_0)$
and the triangle:
$\displaystyle\frac{1}{2}a_0(t-t_0)^2$
To obtain the distance traveled in a time ($\Delta s$) with the position ($s$) and the starting position ($s_0$), resulting in:
| $ \Delta s \equiv s - s_0 $ |
Therefore:
| $ s = s_0 + v_0 ( t - t_0 )+\displaystyle\frac{1}{2} a_0 ( t - t_0 )^2$ |
This corresponds to the general form of a parabola.
ID:(3157, 2)
Path with Constant Acceleration as a Function of the Speed (1)
Equation
In the case of constant acceleration, we can calculate the position ($s$) from the starting position ($s_0$), the initial Speed ($v_0$), the time ($t$), and the start Time ($t_0$) using the equation:
| $ s = s_0 + v_0 ( t - t_0 )+\displaystyle\frac{1}{2} a_0 ( t - t_0 )^2$ |
This allows us to determine the relationship between the distance covered during acceleration/deceleration and the change in velocity:
| $ s = s_1 +\displaystyle\frac{ v_1 ^2- v_{01} ^2}{2 a_1 }$ |
| $ s = s_0 +\displaystyle\frac{ v ^2- v_0 ^2}{2 a_0 }$ |
If we solve for the time ($t$) and the start Time ($t_0$) in the equation of the speed ($v$), which depends on the initial Speed ($v_0$) and the constant Acceleration ($a_0$):
| $ v = v_0 + a_0 ( t - t_0 )$ |
we get:
$t - t_0= \displaystyle\frac{v - v_0}{a_0}$
And when we substitute this into the equation of the position ($s$) with the starting position ($s_0$):
| $ s = s_0 + v_0 ( t - t_0 )+\displaystyle\frac{1}{2} a_0 ( t - t_0 )^2$ |
we obtain an expression for the distance traveled as a function of velocity:
| $ s = s_0 +\displaystyle\frac{ v ^2- v_0 ^2}{2 a_0 }$ |
ID:(3158, 1)
Path with Constant Acceleration as a Function of the Speed (2)
Equation
In the case of constant acceleration, we can calculate the position ($s$) from the starting position ($s_0$), the initial Speed ($v_0$), the time ($t$), and the start Time ($t_0$) using the equation:
| $ s = s_0 + v_0 ( t - t_0 )+\displaystyle\frac{1}{2} a_0 ( t - t_0 )^2$ |
This allows us to determine the relationship between the distance covered during acceleration/deceleration and the change in velocity:
| $ s = s_2 +\displaystyle\frac{ v_2 ^2- v_{02} ^2}{2 a_2 }$ |
| $ s = s_0 +\displaystyle\frac{ v ^2- v_0 ^2}{2 a_0 }$ |
If we solve for the time ($t$) and the start Time ($t_0$) in the equation of the speed ($v$), which depends on the initial Speed ($v_0$) and the constant Acceleration ($a_0$):
| $ v = v_0 + a_0 ( t - t_0 )$ |
we get:
$t - t_0= \displaystyle\frac{v - v_0}{a_0}$
And when we substitute this into the equation of the position ($s$) with the starting position ($s_0$):
| $ s = s_0 + v_0 ( t - t_0 )+\displaystyle\frac{1}{2} a_0 ( t - t_0 )^2$ |
we obtain an expression for the distance traveled as a function of velocity:
| $ s = s_0 +\displaystyle\frac{ v ^2- v_0 ^2}{2 a_0 }$ |
ID:(3158, 2)
Distance traveled (1)
Equation
We can calculate the distance traveled in a time ($\Delta s$) from the starting position ($s_0$) and the position ($s$) using the following equation:
| $ \Delta s_1 \equiv s - s_1 $ |
| $ \Delta s \equiv s - s_0 $ |
ID:(4352, 1)
Distance traveled (2)
Equation
We can calculate the distance traveled in a time ($\Delta s$) from the starting position ($s_0$) and the position ($s$) using the following equation:
| $ \Delta s_2 \equiv s - s_2 $ |
| $ \Delta s \equiv s - s_0 $ |
ID:(4352, 2)