Understand mechanics,from kinematics to oscillations.

Answer

Full solution

1. Differentiate once to get velocity. With $\tau=2\ \mathrm{s}$, $x=(4\ \mathrm{m})e^{-t/\tau}$ gives $v=\dfrac{dx}{dt}=-\dfrac{4}{\tau}e^{-t/\tau}\ \mathrm{m}=-(2\ \mathrm{m/s})\,e^{-t/\tau}$. The exponential is always positive, so $v<0$ for every $t$: the bead moves left the whole time.

2. Differentiate again: $a=\dfrac{dv}{dt}=+\dfrac{2}{\tau}e^{-t/\tau}\ \mathrm{m/s^2}=+(1\ \mathrm{m/s^2})\,e^{-t/\tau}$. This is positive for every $t$.

3. Now use the rule that decides speeding up or slowing down: the speed changes at the rate $\dfrac{d|v|}{dt}=\dfrac{v}{|v|}\,a$, so the speed rises when $v$ and $a$ have the same sign and falls when they have opposite signs. Here $v<0$ and $a>0$, so the speed always falls.

4. Check it directly: the speed is $|v|=2e^{-t/2}$ (in $\mathrm{m/s}$), and $\dfrac{d|v|}{dt}=-e^{-t/2}<0$ for all $t$. The statement is true. Notice the bead never actually stops, because $v$ is never zero, but it slows down forever.

Answer

Full solution

1. Position is the integral of velocity: $x(t)=\int_0^t v_0e^{-s/\tau}\,ds$. An antiderivative of $e^{-s/\tau}$ is $-\tau e^{-s/\tau}$, because the chain rule puts a factor $-\tfrac1\tau$ out front when you differentiate.

2. Evaluate: $x(t)=v_0\left[-\tau e^{-s/\tau}\right]_0^t=v_0\tau\left(1-e^{-t/\tau}\right)$.

3. Take the limit: $e^{-t/\tau}\to0$, so $x\to v_0\tau$. The particle moves forever, since $v$ is never exactly zero, yet it gets only a finite distance because the speed dies off fast enough.

4. Dimension check: $v_0\tau$ has units $\mathrm{(m/s)\cdot s}=\mathrm{m}$, a length, as it must be. Choice (A) has the units of a velocity and choice (B) the units of an acceleration, so neither can be a position. Choice (E) is the position at the single time $t=\tau$, not the limit.