Question

A particle moves along a straight line and its position at time t is given by s(t)=2t3−27t2+84t where s is measured in feet and t in seconds. a) Find the velocity (in ft/sec) of the particle at time t=0: 84 b) The particle stops moving twice, once when t=A and again when t=B where A

Answer 1

a)
`v(0)=84ft/s`

b)
`A=2s`

`B=7s`

c)
`s(t=18s)=4428ft`

d) Δs=4428ft

Step-by-step explanation:

From the exercise we know the equation of position

`s(t)=2t^3-27t^2+84t`

a) To calculate the velocity we need to derivate the equation of position

`v=(ds)/(dt)=6t^2-54t+84`

So, v(0) is:

`v(0)=6(0)^2-54(0)+84=84ft/s`

b) To find the two times where the particle stops A and B we need to solve the quadratic equation:

`0=6t^2-54t+84`

Solving for t

`t=(-b±√(b^2-4ac) )/(2a)`

`a=6\\b=-54\\c=84`

`t=2s` and
`t=7s`

`A=2s\\B=7s`

c)

`s(18s)=2(18)^3-27(18)^2+84(18)=4428ft`

d) Δs=4428ft-0ft=4428ft