Question: A particle moves according to the law of motion
s = t^3 - 6t^2 + 3t, t \geq 0,
where t is measured in seconds and s in feet.
a.) Find the velocity at time t.
b.) What is the velocity after 3 seconds?
c.) When is the particle at rest? Enter your answer as a comma separated list. Enter None if the particle is never at rest.
d.) When is the particle moving in the positive direction?
When 0 \leq t < and t > (Round your answers to 3 decimal places.)
1. **State the problem:**
We have the position function $$s = t^3 - 6t^2 + 3t$$ for $$t \geq 0$$, where $$t$$ is time in seconds and $$s$$ is position in feet.
We need to find:
a) Velocity at time $$t$$.
b) Velocity at $$t=3$$ seconds.
c) Times when the particle is at rest.
d) Intervals when the particle moves in the positive direction.
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2. **Find the velocity function:**
Velocity $$v(t)$$ is the derivative of position $$s(t)$$ with respect to time $$t$$:
$$v(t) = \frac{ds}{dt} = \frac{d}{dt}(t^3 - 6t^2 + 3t)$$
Using power rule:
$$v(t) = 3t^2 - 12t + 3$$
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3. **Calculate velocity at $$t=3$$ seconds:**
Substitute $$t=3$$ into $$v(t)$$:
$$v(3) = 3(3)^2 - 12(3) + 3 = 3(9) - 36 + 3 = 27 - 36 + 3 = -6$$
Rounded to 3 decimal places:
$$v(3) = -6.000$$
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4. **Find when the particle is at rest:**
Particle is at rest when velocity is zero:
$$v(t) = 0 \Rightarrow 3t^2 - 12t + 3 = 0$$
Divide both sides by 3:
$$\cancel{3}t^2 - \cancel{3}4t + \cancel{3} = 0 \Rightarrow t^2 - 4t + 1 = 0$$
Use quadratic formula:
$$t = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(1)}}{2(1)} = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2}$$
Simplify $$\sqrt{12} = 2\sqrt{3}$$:
$$t = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$$
Numerical values:
$$t_1 = 2 - 1.732 = 0.268$$
$$t_2 = 2 + 1.732 = 3.732$$
Since $$t \geq 0$$, both are valid.
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5. **Determine when the particle moves in the positive direction:**
Particle moves positively when velocity $$v(t) > 0$$.
Recall velocity:
$$v(t) = 3t^2 - 12t + 3$$
We found roots at $$t = 0.268$$ and $$t = 3.732$$.
Test intervals:
- For $$0 \leq t < 0.268$$, pick $$t=0$$:
$$v(0) = 3(0)^2 - 12(0) + 3 = 3 > 0$$ (positive)
- For $$0.268 < t < 3.732$$, pick $$t=2$$:
$$v(2) = 3(4) - 12(2) + 3 = 12 - 24 + 3 = -9 < 0$$ (negative)
- For $$t > 3.732$$, pick $$t=4$$:
$$v(4) = 3(16) - 12(4) + 3 = 48 - 48 + 3 = 3 > 0$$ (positive)
So velocity is positive when:
$$0 \leq t < 0.268$$ and $$t > 3.732$$
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**Final answers:**
a) $$v(t) = 3t^2 - 12t + 3$$
b) $$v(3) = -6.000$$
c) Particle at rest at $$t = 0.268, 3.732$$
d) Particle moves positively when $$0 \leq t < 0.268$$ and $$t > 3.732$$