Question

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.

Answer 1

We are given

equation of position

$s(t)=(t)ln(3t)$

Calculation of velocity:

we can find derivative

$s'(t)=1*ln(3t)+t*\frac{3}{3t}$

$s'(t)=ln(3t)+1$

so, velocity is

$v(t)=ln(3t)+1$

now, we can set it to 0

and then we can solve for t

$v(t)=ln(3t)+1=0$

$t=\frac{1}{3e}$

Calculation of acceleration:

we can find derivative again

$v'(t)=\frac{3}{3t} +0$

$v'(t)=\frac{1}{t}$

so, acceleration is

$a(t)=\frac{1}{t}$

now, we can plug value of t

$a(\frac{1}{3e})=\frac{1}{\frac{1}{3e}}$

$a(\frac{1}{3e})=3e$..................Answer