Answer:
a) [tex] v(t) = \frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2}[/tex]
b) 3s
c) t < 3s
d) 1.8ft
e) particle is speeding up when t > 5.385 s and vice versa, slowing down when t < 5.385 s
Explanation:
a)Suppose the equation for motion is:
[tex] f(t) = \frac{9t}{t^2+9} [/tex]
Then the velocity is the derivative of the motion function
[tex] v(t) = \frac{df(t)}{dt} = (9t(t^2+9)^{-1})^'[/tex]
From here we can apply product rule
[tex] v(t) = (9t)^{'}(t^2+9)^{-1} + (9t)((t^2+9)^{-1})^' [/tex]
[tex] v(t) = \frac{9}{t^2+9} - \frac{9t(2t)}{(t^2+9)^2}[/tex]
[tex] v(t) = \frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2}[/tex]
b) The particle is at rest when v(t) = 0:
[tex]\frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2} = 0[/tex]
[tex] \frac{9}{t^2+9} = \frac{18t^2}{(t^2+9)^2}[/tex]
Multiply 2 sides by [tex] (t^2+9)^2 [/tex] we have:
[tex] t^2 + 9 = 2t^2[/tex]
[tex] t^2 = 9[/tex]
[tex] t = 3s [/tex]
(c) The particle is moving in positive direction when v(t) > 0:
[tex]\frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2} > 0[/tex]
[tex] \frac{9}{t^2+9} > \frac{18t^2}{(t^2+9)^2}[/tex]
Multiply 2 sides by [tex] (t^2+9)^2 [/tex] we have:
[tex] t^2 + 9 > 2t^2[/tex]
[tex] t^2 < 9[/tex]
[tex] t < 3s [/tex]
(d) As particle is moving in positive direction when t < 3s and negative direction when t > 3s, we can calculate the distance it's moving up to 3s and then after 3s
[tex]f(3) = \frac{9*3}{3^2+9} = \frac{27}{18} = 1.5 ft[/tex]
At 3s, particle is changing direction to negative, so its position at 6s is
[tex]f(6) = \frac{9*6}{6^2+9} = \frac{54}{45} = 1.2 ft[/tex]
Therefore from 3s to 6s it would have moved a distance of 1.5 - 1.2 = 0.3 ft
Then the total distance it has moved in the first 6 s is 1.5 + 0.3 = 1.8 ft
e) Acceleration is the derivative of velocity function:
[tex]a(t) = \frac{dv(t)}{dt} = (\frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2})^'[/tex]
[tex]a(t) = -\frac{9(2(t^2+9))(2t)}{(t^2+9)^2} - \frac{36t}{(t^2+9)^2} + \frac{18t^2(2(t^2+9))(2t)}{(t^2+9)^3}[/tex]
[tex]a(t) = -\frac{36t}{t^2+9} - \frac{36t}{(t^2+9)^2} + \frac{72t^3}{(t^2+9)^2}[/tex]
[tex]a(t) = -\frac{54t}{t^2+9} + \frac{72t^3 - 36t}{(t^2+9)^2}[/tex]
Particle is speeding up when a(t) > 0:
[tex]-\frac{54t}{t^2+9} + \frac{72t^3 - 36t}{(t^2+9)^2} > 0[/tex]
[tex] \frac{72t^3 - 36t}{(t^2+9)^2} > \frac{54t}{t^2+9}[/tex]
as [tex]t \& (t^2 + 9)^2 \geq 0 [/tex] we can multiply/divide both sides by it:
[tex]8t^2 - 4 > 6(t^2+9) [/tex]
[tex] 8t^2 > 6t^2 + 58 [/tex]
[tex] t^2 > 29 [/tex]
[tex] t > \sqrt(29) \approx 5.385 s [/tex]
so particle is speeding up when t > 5.385 s and vice versa, slowing down when t < 5.385 s
