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

Step-by-step 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=(16\pm √(16^2-4\cdot 3\cdot 2))/(2\cdot 3)`

`t=(16\pm 15.23)/(6)`

`t=5.20 s`