Question

A particle moves according to the law of motion s(t) = t^{3}-8t^{2}+2t, t\ge 0, where t is measured in seconds and s in feet. a.) Find the velocity at time t. Answer: b.) What is the velocity after 3 seconds? Answer: c.) When is the particle at rest? Enter your answer as a comma separated list. Enter None if the particle is never at rest. At t_1= and t_2= with t_1

Answer 1

Answer:

Explanation:

Given

displacement is given by

$s(t)=t^3-8t^2+2t$

so velocity is given by

$v(t)=\frac{\mathrm{d} s(t)}{\mathrm{d} t}$

$v(t)=3t^2-16t+2$

(b)velocity after $t=3 s$

$v(3)=3(3)^2-8\cdot 3+2$

$v(3)=19 m/s$

(c)Particle is at rest

when its velocity will become zero

$v(t)=0$

i.e.  $3t^2-16t+2=0$

$t=\frac{16\pm \sqrt{16^2-4\cdot 3\cdot 2}}{2\cdot 3}$

$t=\frac{16\pm 15.23}{6}$

$t=5.20 s$