Answer:
x(4) ≈ 11.9 m and v(4) ≈ 0.250 m/s
Note: Answers were rounded to three significant figures.
Step-by-step explanation:
To determine the velocity and position of a particle at t = 4 seconds, given its initial velocity, initial position, and a deceleration. We can follow a systematic approach involving integration and differential equations due to the relationship between acceleration, velocity, and position.
We are given:
The acceleration of a partical is the derivative of its velocity with respect to time. Thus, we have:
[tex]a = -2v^3[/tex]
[tex]\Longrightarrow \dfrac{dv}{dt} = -2v^3[/tex]
We can use seperation of variables to solve the above differential equation:
[tex]\boxed{\begin{array}{ccc}\text{\underline{Separable Differential Equation:}}\\\\\displaystyle \frac{dy}{dx} =f(x)g(y)\\\\\displaystyle \rightarrow\int\frac{dy}{g(y)}=\int f(x)dx \end{array}}[/tex]
[tex]\dfrac{dv}{dt} = -2v^3[/tex]
[tex]\Longrightarrow \dfrac{1}{-2v^3}dv = dt[/tex]
[tex]\displaystyle \Longrightarrow \int \dfrac{1}{-2v^3} \, dv = \int dt[/tex]
[tex]\displaystyle \Longrightarrow \dfrac{1}{-2}\int v^{-3} \, dv = t+C[/tex]
Using the power rule for integration:
[tex]\boxed{\begin{array}{ccc}\text{\underline{Power Rule for Integration:}}\\\\\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad \text{for} \quad n \neq -1\end{array}}[/tex]
[tex]\Longrightarrow \dfrac{1}{-2}\left(\dfrac{1}{-2}v^{-2}\right) = t+C[/tex]
[tex]\Longrightarrow \dfrac{1}{4v^2} = t+C[/tex]
Take the above equation and solve for 'v':
[tex]\Longrightarrow \dfrac{1}{v^2} = 4t+ \bar C\\\\\\\\\therefore v = \sqrt{\dfrac{1}{4t+ C}}[/tex]
Using the initial condition, v(0) = 8, we can determine the value of the constant 'C':
[tex]\Longrightarrow 8 = \sqrt{\dfrac{1}{4(0)+ C}}\\\\\\\\[/tex]
[tex]\Longrightarrow 8 = \sqrt{\dfrac{1}{C}}\\\\\\\\[/tex]
[tex]\Longrightarrow 64 = \dfrac{1}{C}\\\\\\\\\therefore C = \dfrac{1}{64}[/tex]
Plug in 'C' into 'v':
[tex]\Longrightarrow v = \sqrt{\dfrac{1}{4t+ \dfrac{1}{64}}}[/tex]
[tex]\therefore v(t) = \dfrac{8}{\sqrt{256t+ 1}}[/tex]
The velocity of a partical is the derivative of its position with respect to time. Thus, we have:
[tex]\dfrac{dx}{dt} = \dfrac{8}{\sqrt{256t+ 1}}[/tex]
Once again, we can use seperation of variables to solve the above differential equation:
[tex]\Longrightarrow dx = \dfrac{8}{\sqrt{256t+ 1}}\, dt[/tex]
[tex]\displaystyle \Longrightarrow \int dx = \int \dfrac{8}{\sqrt{256t+ 1}} \, dt[/tex]
[tex]\displaystyle \Longrightarrow x = 8\int \dfrac{1}{\sqrt{256t+ 1}} \, dt[/tex]
[tex]\displaystyle \Longrightarrow x = 8\int \dfrac{1}{\sqrt{256t+ 1}} \, dt[/tex]
Using u-substitution, let u = 256t + 1, thus du = 256dt.
[tex]\displaystyle \Longrightarrow x = \dfrac{8}{256} \int \dfrac{1}{\sqrt{u}} \, du[/tex]
[tex]\Longrightarrow x = \dfrac{1}{32} \left(2\sqrt{u}+C\right)[/tex]
Plug 'u' back in:
[tex]\Longrightarrow x = \dfrac{1}{32} \left(2\sqrt{256t+1} + C\right)[/tex]
[tex]\therefore x = \dfrac{1}{16}\sqrt{256t+1} + C[/tex]
Using the initial condition, x(0) = 10, we can determine the value of the constant 'C':
[tex]\Longrightarrow 10 = \dfrac{1}{16}\sqrt{256(0)+1} + C[/tex]
[tex]\Longrightarrow 10 = \dfrac{1}{16}\sqrt{1} + C[/tex]
[tex]\Longrightarrow 10 - \dfrac{1}{16} = C\\\\\\\\\therefore C = \dfrac{159}{16}[/tex]
Plug in 'C' into 'x':
[tex]\therefore x(t) = \dfrac{1}{16}\sqrt{256t+1} + \dfrac{159}{16}[/tex]
Thus, we have our position and velocity functions:
[tex]x(t) = \dfrac{1}{16}\sqrt{256t+1} + \dfrac{159}{16} \text{ and } v(t) = \dfrac{8}{\sqrt{256t+ 1}}[/tex]
To determine the position of the particle and its velocity at t = 4, plug in 4 for 't':
[tex]x(4) = \dfrac{1}{16}\sqrt{256(4)+1} + \dfrac{159}{16}\\\\\\\\\therefore x(4) \approx \boxed{11.9 \text{ m}}[/tex]
[tex]v(4) = 8\sqrt{\dfrac{1}{256(4)+ 1}}\\\\\\\\\therefore v(4) \approx \boxed{0.250 \text{ m/s}}[/tex]
Thus, the problem is solved.
