A particle moving along the x axis so that it’s position at t is greater than or equal to 0 is given by s(t)=(t)ln(3t). Find the acceleration of the particle when the velocity is first zero.

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rejkjavik rejkjavik · Answer 1 · 2018-10-13T06:23:25+02:00

We are given

equation of position

[tex]s(t)=(t)ln(3t)[/tex]

Calculation of velocity:

we can find derivative

[tex]s'(t)=1*ln(3t)+t*\frac{3}{3t}[/tex]

[tex]s'(t)=ln(3t)+1[/tex]

so, velocity is

[tex]v(t)=ln(3t)+1[/tex]

now, we can set it to 0

and then we can solve for t

[tex]v(t)=ln(3t)+1=0[/tex]

[tex]t=\frac{1}{3e}[/tex]

Calculation of acceleration:

we can find derivative again

[tex]v'(t)=\frac{3}{3t} +0[/tex]

[tex]v'(t)=\frac{1}{t} [/tex]

so, acceleration is

[tex]a(t)=\frac{1}{t} [/tex]

now, we can plug value of t

[tex]a(\frac{1}{3e})=\frac{1}{\frac{1}{3e}} [/tex]

[tex]a(\frac{1}{3e})=3e [/tex]..................Answer