Compute the velocity function

Let $f(t) = t$ and $g(t) = t^2+1$, so $f'(t) = 1$ and $g'(t) = 2t$. By the quotient rule, $s'(t) = \dfrac{f'(t)g(t) - f(t)g'(t)}{[g(t)]^2} = \dfrac{(1)(t^2+1) - (t)(2t)}{(t^2+1)^2}$.

Simplify the numerator: $(t^2+1) - 2t^2 = 1 - t^2$. Therefore $v(t) = s'(t) = \dfrac{1-t^2}{(t^2+1)^2}$.

Since $(t^2+1)^2 > 0$ for all $t$, the sign of $v(t) = \dfrac{1-t^2}{(t^2+1)^2}$ is determined entirely by the numerator $1-t^2$. Setting $1-t^2 = 0$ gives $t = 1$ (taking only the nonnegative root since $t \ge 0$).

For $0 \le t < 1$, $1-t^2 > 0$, so $v(t) > 0$ and the particle moves right. For $t > 1$, $1-t^2 < 0$, so $v(t) < 0$ and the particle moves left. At $t=1$ the particle is momentarily at rest.

At t equal to two seconds, t is greater than one, so we are in the interval where velocity is negative. From part b, for t greater than one the particle moves left. Therefore just after two seconds the particle is moving left.