Answer
$a_x = 6.42~m/s^2$
Work Step by Step
$v_x(t) = (0.860~m/s^3)~t^2$
We can find the time $t$ when $v_x = 12.0~m/s$
$(0.860~m/s^3)~t^2 = 12.0~m/s$
$t^2 = \frac{12.0~m/s}{0.860~m/s^3}$
$t = \sqrt{\frac{12.0~m/s}{0.860~m/s^3}}$
$t = 3.735~s$
We can use $v_x(t)$ to find an equation for $a_x(t)$.
$a_x(t) = \frac{dv}{dt} = 2(0.860~m/s^3)~t$
$a_x(t) = (1.720~m/s^3)~t$
We can find the acceleration when $t = 3.735~s$.
$a_x = (1.720~m/s^3)(3.735~s) = 6.42~m/s^2$
