The velocity of a particle moving along the x-axis is given, for t0, by v=(50.0t - 2.0t^3) m/s, where t is in seconds. What is the acceleration of the particle when (after t=0) it achieves its maximum displacement in the positive x direction?

Question

The velocity of a particle moving along the x-axis is given, for t>0, by v=(50.0t - 2.0t^3) m/s, where t is in seconds.  What is the acceleration of the particle when (after t=0) it achieves its maximum displacement in the positive x direction?

amistre64 · Answer

since velocity is the derivative of displacement, when v = 0, we have critical points to test for displacement.  the derivative of velocity is acceleration, so when you determine the value of t, plug it in the acceleration function

amistre64 · Answer

$$v(t) = 50t - 2t^3$$
$$2t(25 - t^2)=0$$
so apparently when t= {-5,0,5}; which is only valid for t=5

$$v'(t) = a(t) =50 - 6t^2$$

amistre64 · Answer

something feels wierd there, but i cant see an error in my thought

amistre64 · Answer

if velocity is zero, acceleration would be zero .... is what im usually thinking :/

anonymous · Answer

Thank you for you help!  I haven't been able to check your work against mine yet...but i do appreciate your effort!

amistre64 · Answer

youre welcome.  i do wish i was more confident in the results :)  Im pretty sure it is because we started with a cubic for velocity, or im just missing something.