Answer:
(a) [tex]v(t) = (6.0ti - 21.0t^{2}j - 5.0) m/s[/tex]
(b) [tex]a(t) = (6.0i - 42.0tj) m/s^2[/tex]
(c) [tex]v(2.0) = (12.0i - 84.0j - 5.0) m/s[/tex]
(d) [tex]speed (t=1.0 s) = \sqrt{6^{2}+21^{2}+5^{2}}[/tex] = 22.41 m/s; [tex]speed (t= 3.0 s) = \sqrt{18^{2}+189^{2}+5^{2}}[/tex] = 189.92 m/s
(e) average velocity = (9i-21j-5) m/s
Explanation:
For the given problem:
[tex]r(t) = (3.0t^{2}i - 7.0t^{3}j - 5.0t - 2k) m[/tex]
(a) The velocity as a function of time: v(t) = dr/dt. Thus:
[tex]v(t) = (6.0ti - 21.0t^{2}j - 5.0) m/s[/tex]
(b) The acceleration as a function of time: a(t) = dv/dt. Thus:
[tex]a(t) = (6.0i - 42.0tj) m/s^2[/tex]
(c) The particle velocity at t= 2.0s. Using the equation in part (a);
[tex]v(2.0) = (6.0*2.0i - 21.0(2.0)^{2}j - 5.0) m/s[/tex]
[tex]v(2.0) = (12.0i - 84.0j - 5.0) m/s[/tex]
(d) Its speed at t=1.0s and t=3.0s
[tex]speed = \sqrt{v_{x} ^{2}+v_{y} ^{2}}[/tex]
at t=1.0s
[tex]speed = \sqrt{6^{2}+21^{2}+5^{2}}[/tex] = 22.41 m/s
at t= 3.0s
[tex]speed = \sqrt{18^{2}+189^{2}+5^{2}}[/tex] = 189.92 m/s
(e) The average velocity between t = 1.0 s and t = 2.0 s. Using the equation for r(t).
[tex]r(1.0) = (3.0i - 7.0j - 5.0 - 2k) m[/tex] m
[tex]r(2.0) = (12.0i - 28.0j - 10.0 - 2k) m[/tex]
average velocity = Δr/Δt = (r2-r1)/(t2-t1) = (9i-21j-5) m/s
