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.

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