Question

A particle is moving along the x-axis so that its position at t ≥ 0 is given by s(t)=(t)ln(5t). Find the acceleration of the particle when the velocity is first zero
5e
5e^2
e
None of these

Answer 1

Answer:

$a(\frac{1}{5e})=5e$

Step-by-step explanation:

we are given equation for position function as

$s(t)=tln(5t)$

Since, we have to find acceleration

For finding acceleration , we will find second derivative

$s'(t)=\frac{d}{dt}\left(t\ln \left(5t\right)\right)$

$=\frac{d}{dt}\left(t\right)\ln \left(5t\right)+\frac{d}{dt}\left(\ln \left(5t\right)\right)t$

$=1\cdot \ln \left(5t\right)+\frac{1}{t}t$

$s'(t)=\ln \left(5t\right)+1$

now, we can find derivative again

$s''(t)=\frac{d}{dt}\left(\ln \left(5t\right)+1\right)$

$=\frac{d}{dt}\left(\ln \left(5t\right)\right)+\frac{d}{dt}\left(1\right)$

$=\frac{1}{t}+0$

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

Firstly, we will set velocity =0

and then we can solve for t

$v(t)=s'(t)=\ln \left(5t\right)+1=0$

we get

$t=\frac{1}{5e}$

now, we can plug that into acceleration

and we get

$a(\frac{1}{5e})=\frac{1}{\frac{1}{5e}}$

$a(\frac{1}{5e})=5e$