Answer
(a) $v = \frac{2c^2~t}{2\sqrt{b^2+c^2~t^2}}$
$a =\frac{c^2b^2}{(b^2+c^2~t^2)^{3/2}}$
(b) Since both the velocity and the acceleration are positive for all $t$ such that $t\gt 0$, the particle always moves in the positive direction.
Work Step by Step
(a) $x = \sqrt{b^2+c^2~t^2}$
$v = \frac{dx}{dt}$
We can find the velocity function:
$v = \frac{2c^2~t}{2\sqrt{b^2+c^2~t^2}}$
$a = \frac{dv}{dt}$
We can find the acceleration function:
$a =\frac{(2c^2)~2\sqrt{b^2+c^2~t^2}-(2c^2~t)^2(b^2+c^2~t^2)^{-1/2}}{4(b^2+c^2~t^2)}$
$a =\frac{(2c^2)~2\sqrt{b^2+c^2~t^2}-(2c^2~t)^2(b^2+c^2~t^2)^{-1/2}}{4(b^2+c^2~t^2)}\cdot \frac{\sqrt{b^2+c^2~t^2}}{\sqrt{b^2+c^2~t^2}}$
$a =\frac{(4c^2)(b^2+c^2~t^2)-(2c^2~t)^2}{4(b^2+c^2~t^2)^{3/2}}$
$a =\frac{c^2b^2+c^4~t^2-c^4~t^2}{(b^2+c^2~t^2)^{3/2}}$
$a =\frac{c^2b^2}{(b^2+c^2~t^2)^{3/2}}$
(b) $v = \frac{2c^2~t}{2\sqrt{b^2+c^2~t^2}} \gt 0~~$ when $t\gt 0$
$a =\frac{c^2b^2}{(b^2+c^2~t^2)^{3/2}} \gt 0~~$ for all $t$
Since both the velocity and the acceleration are positive for all $t$ such that $t\gt 0$, the particle always moves in the positive direction.
