9. The displacement of a particle from a fixed source is given as x(t) = t^3 - 2t^2 - 4t + 5 (m). How far is the particle from the source when it goes to momentary rest?
This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Here are the solutions to problems 9 and 10:
9. The displacement of a particle from a fixed source is given as $x(t) = t^3 - 2t^2 - 4t + 5 \text{ (m)}$. How far is the particle from the source when it goes to momentary rest?
To find when the particle is at momentary rest, we first need to find its velocity by differentiating the displacement function with respect to time. Momentary rest occurs when the velocity is zero.
Step 1: Find the velocity function $v(t)$.
$$v(t) = \frac{dx}{dt} = \frac{d}{dt}(t^3 - 2t^2 - 4t + 5)$$
$$v(t) = 3t^2 - 4t - 4 \text{ m/s}$$
Step 2: Set $v(t) = 0$ to find the time(s) when the particle is at rest.
$$3t^2 - 4t - 4 = 0$$
Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$$t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(3)(-4)}}{2(3)}$$
$$t = \frac{4 \pm \sqrt{16 + 48}}{6}$$
$$t = \frac{4 \pm \sqrt{64}}{6}$$
$$t = \frac{4 \pm 8}{6}$$
Two possible values for $t$:
$$t_1 = \frac{4 + 8}{6} = \frac{12}{6} = 2 \text{ s}$$
$$t_2 = \frac{4 - 8}{6} = \frac{-4}{6} = -\frac{2}{3} \text{ s}$$
Since time cannot be negative, we consider $t = 2
