Question

a particle moves along a straight line and its position at time ( t ) is given by ( s(t)=2 t^{3}-21 t^{2}+36 t ) where ( s ) is measured in feet and ( t ) in seconds.
find the velocity (in ( mathrm{ft} / mathrm{sec} ) ) of the particle at time ( t = 0 ):
the particle stops moving (i.e. is in a rest) twice,
first when ( t=),
and again when ( t=)
what is the position of the particle at time 14? ( mathrm{ft} )
finally, what is the total distance the particle travels between time 0 and time 14? ( mathrm{ft} )
question help: message instructor

Explanation:

Step1: Find the velocity function

The velocity function \(v(t)\) is the derivative of the position function \(s(t)\).
Using the power rule \((x^n)^\prime = nx^{n - 1}\), if \(s(t)=2t^{3}-21t^{2}+36t\), then \(v(t)=s^\prime(t)=6t^{2}-42t + 36\).

Step2: Find the velocity at \(t = 0\)

Substitute \(t = 0\) into \(v(t)\):
\(v(0)=6(0)^{2}-42(0)+36=36\).

Step3: Find when the particle stops moving

Set \(v(t)=0\), so \(6t^{2}-42t + 36 = 0\).
Divide through by \(6\): \(t^{2}-7t + 6=0\).
Factor: \((t - 1)(t - 6)=0\).
Using the zero - product property \(t-1 = 0\) or \(t - 6=0\), so \(t = 1\) or \(t = 6\).

Step4: Find the position at \(t = 14\)

Substitute \(t = 14\) into \(s(t)\):
\(s(14)=2(14)^{3}-21(14)^{2}+36(14)\)
\(=2\times2744-21\times196 + 36\times14\)
\(=5488-4116+504\)
\(=1876\).

Step5: Find the total distance

We need to consider the intervals \([0,1]\), \([1,6]\), and \([6,14]\).

• For \(t\in[0,1]\): \(v(t)=6t^{2}-42t + 36\), \(v(t)>0\) (test \(t = 0.5\), \(v(0.5)=6\times(0.5)^{2}-42\times(0.5)+36=1.5-21 + 36=16.5>0\))

\(s(1)-s(0)=(2\times1^{3}-21\times1^{2}+36\times1)-(2\times0^{3}-21\times0^{2}+36\times0)=2 - 21+36=17\)

• For \(t\in[1,6]\): \(v(t)=6t^{2}-42t + 36\), \(v(t)<0\) (test \(t = 2\), \(v(2)=6\times2^{2}-42\times2+36=24-84 + 36=-24<0\))

\(s(6)-s(1)=(2\times6^{3}-21\times6^{2}+36\times6)-(17)=(432-756 + 216)-17=-125\) (distance is \(|s(6)-s(1)| = 125\))

• For \(t\in[6,14]\): \(v(t)=6t^{2}-42t + 36\), \(v(t)>0\) (test \(t = 7\), \(v(7)=6\times7^{2}-42\times7+36=294-294+36=36>0\))

\(s(14)-s(6)=(1876)-(2\times6^{3}-21\times6^{2}+36\times6)=1876-(432-756 + 216)=1876 + 108=1984\)

The total distance \(D=17 + 125+1984=2126\).

Answer:

The velocity at \(t = 0\) is \(36\) ft/sec.
The particle stops at \(t = 1\) and \(t = 6\).
The position at \(t = 14\) is \(1876\) ft.
The total distance is \(2126\) ft.