Answer
(a) $0.077s$
(b) $0.46~μm$
Work Step by Step
(a) We know that
$v_f=v_{\circ}+at$
We plug in the known values to obtain:
$12\mu m/s=0=(156\mu m/s)t$
This simplifies to:
$t=0.077s$
(b) The required distance can be determined as:
$x_{t}=x_{\circ}+v_{\circ}t+\frac{1}{2}at^2$
We plug in the known values to obtain:
$x_{t}=0+0+\frac{1}{2}(156\mu m/s^2)(0.077s^2)$
$x_{t}=0.46\mu m$
