Respuesta :
Answer:
(a). The velocity is 0.099 m/s.
(b). The position is 19.75 m.
Explanation:
Given that,
The deceleration is
[tex]a=(-2v^3)\ m/s^2[/tex]
We need to calculate the velocity at t = 25 s
The acceleration is the first derivative of velocity of the particle.
[tex]a=\dfrac{dv}{dt}[/tex]
[tex]\dfrac{dv}{dt}=-2v^3[/tex]
[tex]\dfrac{dv}{-v^3}=2dt[/tex]
On integrating
[tex]int{\dfrac{dv}{-v^3}}=\int{2dt}[/tex]
[tex]\dfrac{1}{2v^2}=2t+C[/tex]
[tex]v^2=\dfrac{1}{4t+2C}[/tex]....(I)
At t = 0, v = 10 m/s
[tex]10^2=\dfrac{1}{4\times0+2C}[/tex]
[tex]C=\dfrac{1}{200}[/tex]
Put the value of C in equation (I)
[tex]v^2=\dfrac{1}{4\times25+2\times\dfrac{1}{200}}[/tex]
[tex]v=\sqrt{\dfrac{1}{4\times25+2\times\dfrac{1}{200}}}[/tex]
[tex]v=0.099\ m/s[/tex]
The velocity is 0.099 m/s.
(b). We need to calculate the position at t = 25 sec
The velocity is the first derivative of position of the particle.
[tex]\dfrac{ds}{dt}=v[/tex]
On integrating
[tex]\int{ds}=\int(\sqrt{\dfrac{200}{800t+1}})dt[/tex]
[tex]s=\dfrac{\sqrt{200}\times2\sqrt{800t+1}}{800}+C'[/tex]
At t = 0, s = 15 m
[tex]15=\dfrac{200}{800}+C'[/tex]
[tex]C'=15-\dfrac{200}{800}[/tex]
[tex]C'=14.75[/tex]
Put the value in the equation
[tex]s=\dfrac{\sqrt{200}\times2\sqrt{800\times25+1}}{800}+14.75[/tex]
[tex]s=19.75\ m[/tex]
The position is 19.75 m.
Hence, (a). The velocity is 0.099 m/s.
(b). The position is 19.75 m.
The velocity of an object depends on its position.
Part A: The velocity of the moving particle at t = 25 s is 0.099 m/s.
Part B: The position of the particle at t = 25 s is 15.475 m.
What is velocity?
Velocity is defined as the rate at which the position of an object changes.
Given that the deacceleration of the particle is [tex]a = (-2v^3)[/tex]. The velocity of the particle v = 10 m/s and a position s = 15 m when t = 0.
The acceleration of the particle is given below.
[tex]a = \dfrac {dv}{dt}[/tex]
[tex]-2v^3 = \dfrac {dv}{dt}[/tex]
[tex]- 2 dt = \dfrac {1}{v^3}dv[/tex]
Integrating the above equation, we get,
[tex]\int (-2 dt) = \int \dfrac {1}{v^3} dv[/tex]
[tex]2t + C = \dfrac {1}{2v^2}[/tex]
[tex]v^2 = \dfrac {1}{4t + 2C}[/tex]
Putting t=0 s, v = 10 m/s
[tex]100 = \dfrac {1}{4\times 0 + 2C}[/tex]
[tex]100 = \dfrac {1}{2C}[/tex]
[tex]C = 5 \times 10^{-3}[/tex]
Part A: Velocity
The velocity of the particle at t = 25 s is given as below.
[tex]v^2 = \dfrac { 1}{(4\times 25) + (2\times 5\times 10^{-3})}[/tex]
[tex]v^2 = \dfrac{1}{100.01}[/tex]
[tex]v = 0.099 \;\rm m/s[/tex]
Hence the velocity of the moving particle at t = 25 s is 0.099 m/s.
Part B: Position
The velocity of the particle is given as,
[tex]v = \dfrac {ds}{dt}[/tex]
[tex]v dt = ds[/tex]
Integrating the above equation, we get,
[tex]\int vdt = \int ds[/tex]
[tex]\int \sqrt{\dfrac {1}{4t + (\dfrac {1}{100})}} dt = s[/tex]
[tex]\int\sqrt{ {\dfrac {100}{400 t + 1}}}dt = s[/tex]
[tex]s =\dfrac {\sqrt{100} \times 2 \sqrt{400 t + 1}}{400} + C'[/tex]
Put t = 0 s, and s = 10 m,
[tex]15=\dfrac {\sqrt{100} \times 2 \sqrt{400 \times 0 + 1}}{400} + C'[/tex]
[tex]15 = \dfrac {\sqrt{100}\times 2}{400} + C'[/tex]
[tex]C' = 14.95[/tex]
Now the position at t = 25 s is,
[tex]s =\dfrac {\sqrt{100} \times 2 \sqrt{400 \times 25 + 1}}{400} + 14.95[/tex]
[tex]s =15.475 \;\rm m[/tex]
Hence the position of the particle at t = 25 s is 15.475 m.
To know more about the velocity and position, follow the link given below.
https://brainly.com/question/1439172.
