The position of a particle at any time t, t is (greater than or equal to) 0 is given by s(t)=ln(t+1)+1/8t^2-4t. In which direction is the particle moving when the acceleration is zero?

Question

anonymous · Answer

Did you mean to write s(t)=...?

Construct the equations for velocity and acceleration.

v(t)=s'(t)
a(t)=v'(t)

Then set a(t)=0 and solve for t. Plug this value of t into v(t) to see which way it's going :)

lillian_a · Answer

$$ s(t)=\ln (t+1)+1/8t^2-4t$$

anonymous · Answer

Can you tell me what v(t) and a(t) are now?

anonymous · Answer

I've got to go now so I'm just posting the answer so you can check yours when you're done.
$$s(t)=ln(t+1)+\dfrac{1}{8t^2}-4t$$$$v(t)=s'(t)=\dfrac{1}{t+1}-\dfrac{8}{t^2}-4$$$$a(t)=v'(t)=-\dfrac{1}{(t+1)^2} +\dfrac{16}{t^3} $$
$$0=-\dfrac{1}{(t+1)^2} +\dfrac{16}{t^3}=-t^3+16(t+1)^2=-t^3+16t^2+32t+16$$$$t\approx17.844$$
$$v(17.844)=-3.972057681$$So the particle is moving backwards. I think this is right, but it's been a while since I've done any mechanics.