The acceleration of the particle in any time will be the second derivative of s(t)=(t)ln (2t)
1)We have to find the first derivative.
v(t)=velocitiy
v(t)=ds/dt=ln (2t)+2t/2t
v(t)=ds/dt=ln (2t)+1
2) we find out the value of "t" when v(t)=0
ln (2t)+1=0
ln (2t)=-1 ⇔ e⁻¹=2t ⇒t=1/2e
3)We have to find the second derivative of s(t)
a(t)= acceleration
a(t)=d²/d²t=2/2t+0=1/t
3) We find the acceleration of the particle at t=1/2e
a(1/2e)=1/(1/2e)=2e
Answer: the acceleration of the particle when the veolocity is zero would be 2e u/t² (u=units of lenght; t=units of time).
Answer: 2e