The function v(t) is the velocity in m/sec of a particle moving along the x-axis. Use analytic methods to do each of the following: (a) Determine when the particle is moving to the right, to the left, and stopped. (b) Find the particle's displacement for the given time interval. If s(0) = 3, what is the particle's final position? (c) Find the total distance traveled by the particle. v(t) = 5 (sint)^2(cost); 0 ≤ t ≤ 2π

Answer:

(a) The particle is moving to the right in the interval  [tex](0 \ , \ \displaystyle\frac{\pi}{2}) \ \cup \ (\displaystyle\frac{3\pi}{2} \ , \ 2\pi)[/tex] , to the left in the interval [tex](\displaystyle\frac{\pi}{2}\ , \ \displaystyle\frac{3\pi}{2})[/tex], and stops when t = 0, [tex]\displaystyle\frac{\pi}{2}[/tex], [tex]\displaystyle\frac{3\pi}{2}[/tex] and [tex]2\pi[/tex].

(b) The equation of the particle's displacement is [tex]\mathrm{s(t)} \ = \ \displaystyle\frac{5}{3} \ \mathrm{sin^{3}(t)} \ + \ 3[/tex]; Final position of the particle [tex]\mathrm{s(2\pi)} \ = \ 3[/tex].

(c)  The total distance traveled by the particle is 9.67 (2 d.p.)

Step-by-step explanation:

(a) The particle is moving towards the right direction when v(t) > 0 and to the left direction when v(t) < 0. It stops when v(t) = 0 (no velocity).

Situation 1: When the particle stops.

[tex]\-\hspace{1.7cm} v(t) \ = \ 0 \\ \\ 5 \ \mathrm{sin^{2}(t)} \ \mathrm{cos(t)} \ = \ 0 \\ \\ \-\hspace{0.3cm} \mathrm{sin^{2}(t) \ cos(t)} \ = \ 0 \\ \\ \mathrm{sin^{2}(t)} \ = \ 0 \ \ \ \mathrm{or} \ \ \ \mathrm{cos(t)} \ = \ 0 \\ \\ \-\hspace{0.85cm} t \ = \ 0, \ \displaystyle\frac{\pi}{2}, \ \displaystyle\frac{3\pi}{2} \ \ \mathrm{and} \ \ 2\pi[/tex].

Situation 2: When the particle moves to the right.

[tex]\-\hspace{1.67cm} v(t) \ > \ 0 \\ \\ 5 \ \mathrm{sin^2(t) \ cos(t)} \ > \ 0[/tex]

Since the term [tex]5 \ \mathrm{sin^{2}(t)}[/tex] is always positive for all value of t of the interval [tex]0 \ \leq \mathrm{t} \leq \ 2\pi[/tex], hence the determining factor is cos(t). Then, the question becomes of when is cos(t) positive? The term cos(t) is positive in the first and third quadrant or when [tex]\mathrm{t} \ \epsilon \ (0, \ \displaystyle\frac{\pi}{2}) \ \cup \ (\displaystyle\frac{3\pi}{2}, \ 2\pi)[/tex] .  

*Note that parentheses are used to demonstrate the interval of t in which cos(t) is strictly positive, implying that the endpoints of the interval are non-inclusive for the set of values for t.

Situation 3: When the particle moves to the left.

[tex]\-\hspace{1.67cm} v(t) \ < \ 0 \\ \\ 5 \ \mathrm{sin^2(t) \ cos(t)} \ < \ 0[/tex]

Similarly, the term [tex]5 \ \mathrm{sin^{2}(t)}[/tex] is always positive for all value of t of the interval [tex]0 \ \leq \mathrm{t} \leq \ 2\pi[/tex], hence the determining factor is cos(t). Then, the question becomes of when is cos(t) positive? The term cos(t) is negative in the second and third quadrant or  [tex]\mathrm{t} \ \epsilon \ (\displaystyle\frac{\pi}{2}, \ \displaystyle\frac{3\pi}{2})[/tex].

(b) The equation of the particle's displacement can be evaluated by integrating the equation of the particle's velocity.

[tex]s(t) \ = \ \displaystyle\int\ {5 \ \mathrm{sin^{2}(t) \ cos(t)}} \, dx \ \\ \\ \-\hspace{0.69cm} = \ 5 \ \displaystyle\int\ \mathrm{sin^{2}(t) \ cos(t)} \, dx[/tex]

To integrate the expression [tex]\mathrm{sin^{2}(t) \ cos(t)}[/tex], u-substitution is performed where

[tex]u \ = \ \mathrm{sin(t)} \ , \ \ du \ = \ \mathrm{cos(t)} \, dx[/tex].

[tex]s(t) \ = \ 5 \ \displaystyle\int\ \mathrm{sin^{2}(t) \ cos(t)} \, dx \\ \\ \-\hspace{0.7cm} = \ 5 \ \displaystyle\int\ \ \mathrm{sin^{2}(t)} \, du \\ \\ \-\hspace{0.7cm} = \ 5 \ \displaystyle\int\ \ u^{2} \, du \\ \\ \-\hspace{0.7cm} = \ \displaystyle\frac{5u^{3}}{3} \ + \ C \\ \\ \-\hspace{0.7cm} = \ \displaystyle\frac{5}{3} \ \mathrm{sin^{3}(t)} \ + \ C \\ \\ s(0) \ = \ \displaystyle\frac{5}{3} \ \mathrm{sin^{3}(0)} \ + \ C \\ \\ \-\hspace{0.48cm} 3 \ = \ 0 \ + \ C \\ \\ \-\hspace{0.4cm} C \ = \ 3.[/tex]

Therefore, [tex]s(t) \ = \ \displaystyle\frac{5}{3} \ \mathrm{sin^{3}(t)} \ + \ 3[/tex].

The final position of the particle is [tex]s(2\pi) \ = \ \displaystyle\frac{5}{3} \ \mathrm{sin^{3}(2\pi)} \ + \ 3 \ = \ 3[/tex].

(c)

[tex]s(\displaystyle\frac{\pi}{2}) \ = \ \displaystyle\frac{5}{3} \ \mathrm{sin^{3}(\frac{\pi}{2})} \ + \ 3 \\ \\ \-\hspace{0.85cm} \ = \ \displaystyle\frac{14}{3} \qquad (\mathrm{The \ distance \ traveled \ initially \ when \ moving \ to \ the \ right})[/tex]

[tex]|s(\displaystyle\frac{3\pi}{2}) - s(\displatstyle\frac{\pi}{2})| \ = \ |\displaystyle\frac{5}{3} \ (\mathrm{sin^{3}(\frac{3\pi}{2})} \ - \ \mathrm{sin^{3}(\displaystyle\frac{\pi}{2})})| \ \\ \\ \-\hspace{2.28cm} \ = \ \displaystyle\frac{5}{3} | (-1) \ - \ 1| \\ \\ \-\hspace{2.42cm} = \displaystyle\frac{10}{3} \\ \\ (\mathrm{The \ distance \ traveled \ when \ moving \ to \ the \ left})[/tex]

[tex]|s(2\pi) - s(\displaystyle\frac{3\pi}{2})| \ = \ |\displaystyle\frac{5}{3} \ (\mathrm{sin^{3}(2\pi})} \ - \ \mathrm{sin^{3}(\displaystyle\frac{3\pi}{2})})| \ \\ \\ \-\hspace{2.28cm} \ = \ \displaystyle\frac{5}{3} | 0 \ - \ 1| \\ \\ \-\hspace{2.42cm} = \displaystyle\frac{5}{3} \\ \\ (\mathrm{The \ distance \ traveled \ finally \ when \ moving \ to \ the \ right})[/tex].

The total distance traveled by the particle in the given time interval is[tex]\displaystyle\frac{14}{3} \ + \ \displaystyle\frac{5}{3} \ + \ \displaystyle\frac{10}{3} \ = \ \displaystyle\frac{29}{3}[/tex].

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