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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

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amhugueth

Answer:

a) [tex]v(0)=84ft/s[/tex]

b) [tex]A=2s[/tex]

[tex]B=7s[/tex]

c) [tex]s(t=18s)=4428ft[/tex]

d) Δs=4428ft

Explanation:

From the exercise we know the equation of position

[tex]s(t)=2t^3-27t^2+84t[/tex]

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

[tex]v=\frac{ds}{dt}=6t^2-54t+84[/tex]

So, v(0) is:

[tex]v(0)=6(0)^2-54(0)+84=84ft/s[/tex]

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

[tex]0=6t^2-54t+84[/tex]

Solving for t

[tex]t=\frac{-b±\sqrt{b^2-4ac} }{2a}[/tex]

[tex]a=6\\b=-54\\c=84[/tex]

[tex]t=2s[/tex] and [tex]t=7s[/tex]

[tex]A=2s\\B=7s[/tex]

c)

[tex]s(18s)=2(18)^3-27(18)^2+84(18)=4428ft[/tex]

d) Δs=4428ft-0ft=4428ft

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