Question

a particle moving along the x axis has a position given by x = 54t - 2.0t³ m. at the time t = 3.0 s, the speed of the particle is zero. which statement is correct? a. the particle remains at rest after t = 3.0 s. b. the particle no longer accelerates after t = 3.0 s. c. the particle can be found at positions x < 0 m only when t < 0 s. d. all of the above are correct. e. none of the above is correct.

Explanation:

Step1: Find the velocity function

The velocity $v$ is the derivative of the position function $x(t)$. Given $x = 54t-2.0t^{3}$, using the power - rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we have $v=\frac{dx}{dt}=54 - 6t^{2}$.

Step2: Find the acceleration function

The acceleration $a$ is the derivative of the velocity function. So, $a=\frac{dv}{dt}=- 12t$.

Step3: Analyze option a

At $t = 3.0\ s$, $v = 0$. But since $a=-12t$, when $t>3\ s$, $a
eq0$ (for example, at $t = 4\ s$, $a=-48\ m/s^{2}$), so the particle will start moving again after $t = 3\ s$. Option a is incorrect.

Step4: Analyze option b

Since $a=-12t$, for $t>3\ s$, $a
eq0$. The particle still accelerates after $t = 3\ s$. Option b is incorrect.

Step5: Analyze option c

Set $x=54t - 2t^{3}<0$. We can factor out $2t$ to get $2t(27 - t^{2})<0$, or $2t(3\sqrt{3}-t)(3\sqrt{3}+t)<0$. The roots of the equation $x = 0$ are $t = 0,t = 3\sqrt{3}\ s,t=-3\sqrt{3}\ s$. By testing intervals, we find that $x<0$ for $t>3\sqrt{3}\ s$ as well. Option c is incorrect.

Answer:

E. None of the above is correct.