Answer
a) $v=\frac{c^{2}t}{\sqrt {b^{2}+c^{2}t^{2}}}$; $a=\frac{c^{2}b^{2}}{(b^{2}+c^{2}t^{2})\sqrt {b^{2}+c^{2}t^{2}}}$
b) For all $t\geq 0$, $v\geq 0$
Work Step by Step
a) The velocity is:
$$v=x'=\frac{(b^{2}+c^{2}t^{2})'}{2\sqrt {b^{2}+c^{2}t^{2}}}=\frac{2c^{2}t}{2\sqrt {b^{2}+c^{2}t^{2}}}=\frac{c^{2}t}{\sqrt {b^{2}+c^{2}t^{2}}}$$
The acceleration is:
$$a=v'=\frac{(c^{2}t)'\sqrt {b^{2}+c^{2}t^{2}}-c^{2}t(\sqrt {b^{2}+c^{2}t^{2}})'}{(\sqrt {b^{2}+c^{2}t^{2}})^{2}}=\frac{c^{2}\sqrt {b^{2}+c^{2}t^{2}}-c^{2}t\frac{c^{2}t}{\sqrt {b^{2}+c^{2}t^{2}}}}{b^{2}+c^{2}t^{2}}=\frac{c^{2}b^{2}}{(b^{2}+c^{2}t^{2})\sqrt {b^{2}+c^{2}t^{2}}}$$
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b) For all $t\geq 0$, the velocity $v=\frac{c^{2}t}{\sqrt {b^{2}+c^{2}t^{2}}}$ is always positive equivalent to say that the rate of change of $x$ with respect to $t$ is always positive which means that the variation of $x$ is always positive as $t$ increase so the object always moves in the $+$ direction.
