Answer to Question #85226 in Mechanics | Relativity for Aileen Erika R. Elmido

Question

The coordinate of a particle in meters is given by x(t) = 16t - 3.0t^3, where the time t is in seconds. The particle is momentarily at rest at t=?

Answer 1

The particle is momentarily at rest when its speed is 0. How to find the speed? Derive it by taking a derivative of coordinate:

$$v(t)=x'(t)=(16t - 3.0t^3)'=16-3\cdot 3.0t^{3-1}=0,$$

hence:

$$0=16-9t^2,$$ $$t=\sqrt{16/9}=4/3\text{ s}.$$