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A particle, initially at rest, moves along the $x$-axis so that its acceleration at any time $t \geq 0$ is given by $a(t)=12 t^{2}-4$. The position of the particle when $t=1$ is $\boldsymbol{x}(1)=3$.
(a) Find the values of $t$ for which the particle is at rest.
(b) Write an expression for the position $\boldsymbol{x}(\boldsymbol{l})$ of the particle at any time $t \geq 0$.
(c) Find the total distance traveled by the particle from $t=0$ to $t=2$.
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Answer from Sia

Posted about 1 month ago

Solution by Steps

step 1

The particle is at rest when its velocity $v(t) = 0$. To find $v(t)$, we integrate the acceleration function $a(t) = 12t^2 - 4$

step 2

Integrate $a(t)$: $\int (12t^2 - 4) \, dt = 4t^3 - 4t + C$

step 3

Since the particle is initially at rest, $v(0) = 0$. Therefore, $0 = 4(0)^3 - 4(0) + C \implies C = 0$. Thus, $v(t) = 4t^3 - 4t$

step 4

Set $v(t) = 0$: $4t^3 - 4t = 0 \implies 4t(t^2 - 1) = 0 \implies t(t - 1)(t + 1) = 0$

step 5

Solve for $t$: $t = 0, t = 1, t = -1$. Since $t \geq 0$, the values are $t = 0$ and $t = 1$

Answer

The particle is at rest at $t = 0$ and $t = 1$.

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(b) Write an expression for the position $\boldsymbol{x}(t)$ of the particle at any time $t \geq 0$.
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step 1

To find the position $x(t)$, we integrate the velocity function $v(t) = 4t^3 - 4t$

step 2

Integrate $v(t)$: $\int (4t^3 - 4t) \, dt = t^4 - 2t^2 + C$

step 3

Use the initial condition $x(1) = 3$ to find $C$: $3 = (1)^4 - 2(1)^2 + C \implies 3 = 1 - 2 + C \implies C = 4$

step 4

Therefore, the position function is $x(t) = t^4 - 2t^2 + 4$

Answer

The position of the particle at any time $t \geq 0$ is $x(t) = t^4 - 2t^2 + 4$.

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(c) Find the total distance traveled by the particle from $t=0$ to $t=2$.
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step 1

The total distance traveled is the integral of the absolute value of the velocity function $v(t) = 4t^3 - 4t$ from $t=0$ to $t=2$

step 2

Find the points where $v(t) = 0$ within the interval $[0, 2]$: $4t(t^2 - 1) = 0 \implies t = 0, t = 1$

step 3

Split the integral at these points: $\int_0^2 |4t^3 - 4t| \, dt = \int_0^1 |4t^3 - 4t| \, dt + \int_1^2 |4t^3 - 4t| \, dt$

step 4

Evaluate the integrals:
- For $0 \leq t \leq 1$, $4t^3 - 4t \leq 0$, so $|4t^3 - 4t| = -(4t^3 - 4t) = 4t - 4t^3$.
- For $1 \leq t \leq 2$, $4t^3 - 4t \geq 0$, so $|4t^3 - 4t| = 4t^3 - 4t$

step 5

Integrate each part:
$\int_0^1 (4t - 4t^3) \, dt = \left[2t^2 - t^4\right]_0^1 = 2 - 1 = 1$,
$\int_1^2 (4t^3 - 4t) \, dt = \left[t^4 - 2t^2\right]_1^2 = (16 - 8) - (1 - 2) = 8 + 1 = 9$

step 6

Add the results: $1 + 9 = 10$

Answer

The total distance traveled by the particle from $t=0$ to $t=2$ is $10$ units.

Key Concept

Integration of acceleration and velocity functions

Explanation

To find when the particle is at rest, we integrate the acceleration to get velocity and solve for when it is zero. To find the position, we integrate the velocity and use initial conditions. To find the total distance, we integrate the absolute value of the velocity over the given interval.

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