Question

The position of a particle for t > 0 is given by ????⃗ (????)=(m.

Answer 1

Answer:

(a) $v(t) = (6.0ti - 21.0t^{2}j - 5.0) m/s$

(b) $a(t) = (6.0i - 42.0tj) m/s^2$

(c) $v(2.0) = (12.0i - 84.0j - 5.0) m/s$

(d) $speed (t=1.0 s) = \sqrt{6^{2}+21^{2}+5^{2}}$ = 22.41 m/s; $speed (t= 3.0 s) = \sqrt{18^{2}+189^{2}+5^{2}}$ = 189.92 m/s

(e) average velocity = (9i-21j-5) m/s

Explanation:

For the given problem:

$r(t) = (3.0t^{2}i - 7.0t^{3}j - 5.0t - 2k) m$

(a) The velocity as a function of time: v(t) = dr/dt. Thus:

$v(t) = (6.0ti - 21.0t^{2}j - 5.0) m/s$

(b) The acceleration as a function of time: a(t) = dv/dt. Thus:

$a(t) = (6.0i - 42.0tj) m/s^2$

(c) The particle velocity at t= 2.0s. Using the equation in part (a);

$v(2.0) = (6.0*2.0i - 21.0(2.0)^{2}j - 5.0) m/s$

$v(2.0) = (12.0i - 84.0j - 5.0) m/s$  

(d) Its speed at t=1.0s and t=3.0s

$speed = \sqrt{v_{x} ^{2}+v_{y} ^{2}}$

at t=1.0s

$speed = \sqrt{6^{2}+21^{2}+5^{2}}$ = 22.41 m/s

at t= 3.0s

$speed = \sqrt{18^{2}+189^{2}+5^{2}}$ = 189.92 m/s

(e)  The average velocity between t = 1.0 s and t = 2.0 s. Using the equation for r(t).

$r(1.0) = (3.0i - 7.0j - 5.0 - 2k) m$ m

$r(2.0) = (12.0i - 28.0j - 10.0 - 2k) m$

average velocity = Δr/Δt = (r2-r1)/(t2-t1) = (9i-21j-5) m/s