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a particle is moving along the x axis so that its position at time t>/= 0 is given by s(t)=(t)ln(2t). Find the acceleration of the particle when the velocity is first zero.

Respuesta :

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

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