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nikitadnepr [17]
1 year ago
5

The displacement of a particle from a reference point at any instant is given by s = 3 t 2 - 4 t + 5 , where s is in metres and

t is time in seconds. calculate the average velocity of the particle in the time interval between 3 s and 5 s, and its instantaneous velocity at 4 s.
Mathematics
1 answer:
lina2011 [118]1 year ago
3 0

The average velocity of the particle in the time interval between 3s and 5s is 20 ms⁻¹ and its instantaneous velocity at 4s is 20 ms⁻¹.

How to determine average velocity and instantaneous velocity?
Average velocity is defined as the body's overall displacement divided by its time of motion. While instantaneous velocity is defined as a body's speed at a certain instant in time, or its displacement at that instant. When the velocity is constant, average and instantaneous velocities will equalize at just one condition.
The definition of instantaneous velocity is the rate of change of position over a relatively brief time period (almost zero). Simply said, the speed of an object at that precise moment. The definition of instantaneous velocity is "The velocity of an item in motion at a certain point in time." The instantaneous velocity of an object may be equal to its standard velocity if it has uniform velocity.
Mathematically, average velocity = [s(t₂) - s(t₁)]/[t₂ - t₁]
Instantaneous velocity at time, t is = (ds/dt) at time = t

Given, the displacement for the particle is given by s = 3t² - 4t + 5
Time interval, t₁ = 5s and t₂ = 3s;
Using formula in literature, average velocity of the particle in the time interval between 3s and 5s is:
Average velocity = (s(5) - s(3))/(5 - 3) = (60 - 20)/2 = 20 ms⁻¹
Instantaneous velocity at t = 4 is ds/dt at that time-frame:
Now, v = ds/dt = 6t -4
Now, v(4) = 6(4) - 4 = 20 ms⁻¹
The average velocity of the particle in the time interval between 3s and 5s is 20 ms⁻¹ and its instantaneous velocity at 4s is 20 ms⁻¹.

To learn more about this, tap on the link below:
brainly.com/question/2234298

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The table shows the number of games a chess player won in professional competitions, based on the number of games played.
AfilCa [17]

slope  = (25-4)/(30-10)

 slope = 21/20

slope =21/20  

using point slope form

y-y1 = m(x-x1)

y-4 = 21/20 (x-10)

y = 21/20x -21/2 +4

y = 21/20 x -21/2 +8/2

y = 21/20x -13/2


let x=40

y = 21/20 (40) -13/2

y = 42-13/2

y = 35.5 games


If we round the slope to 1

slope =1

using point slope form

y-y1 = m(x-x1)

y-4 = 1 (x-10)

y = 1x -10 +4

y =  x -6


let x=40

y = 40-6

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A matea le encanta dibujar barquitos.Si traslado su barquito inicial 1 des veces,(2 y 3) ¿Cómo desplazo cada uno de sus puntos?
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Answer:

Luego de una búsqueda online, encuentre más información sobre esta pregunta, la cual se puede ver en la imagen del final.

Básicamente lo que tenemos que hacer en este tipo de problemas, es estudiar como los puntos del barco uno cambian cuando hacemos cada transformación, usualmente con un par de puntos ya podremos notar un patrón, y de esta forma entender la transformación que utilizo para mover la figura.

Para conocer las coordenadas de cada punto tenemos que ver el valor de x asociado (el eje horizontal) y el valor de y asociado (el eje horizontal), notar que cada punto está representado por una letra, esto nos permite asociar a los puntos de los distintos barcos.

Entonces, en la transformación de (1) a (2), los cambios en los puntos son:

A (1, 7) --> A₁ (11, 11)                (sumamos 10 al eje x, sumamos 4 al eje y)

B (0, 9) --> B₁ (10, 13)             (sumamos 10 al eje x, sumamos 4 al eje y)

C (3 , 9) --> C₁ (13, 13)             (sumamos 10 al eje x, sumamos 4 al eje y)

Ok, ya con 3 puntos podemos notar un patrón, sea (x, y) un punto cualquiera en el barco (1), el correspondiente punto en el barco (2) va a ser (x + 10, y  + 4), esta es la regla utilizada para mover el barco de (1) a (2), lo que es equivalente a decir que movió el barco 10 unidades a la derecha, y 4 unidades para arriba.

Ahora Mateo vuelve a mover el barco, esta vez de (2) a (3).

De vuelta, tenemos que ver como los puntos del barco (2) cambian, y tratar de encontrar un patrón:

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B₁ (10, 13) --> B₂ (8, 3)   (restamos 2 en el eje x, restamos 10 en el eje y)

C₁ (13, 13) --> C₂ (11, 3)   (restamos 2 en el eje x, restamos 10 en el eje y)

De vuelta, ya tenemos suficiente información como para notar el patrón, para un punto (x, y) del barco (2), el correspondiente punto en el barco (3) se escribe como (x - 2, y - 10).

Es decir, movemos el barco (2) 2 unidades para la izquierda y 10 unidades para abajo, asi llegando al barco (3).

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