Answer
(a) $21~m$
(b) Greater than 6.0 m/s.
$v=8.6~m/s$
Work Step by Step
(a) We know that
$v_f^2=v_i^2+2a\Delta x$
This can be rearranged as:
$\Delta x=\frac{v_i^2}{2a}$
We plug in the known values to obtain:
$\Delta x=\frac{-(12m/s)^2}{2(-3.5m/s^2)}=21m$
(b) We know that
$v_f=[v_i^2+2a\Delta x]^{\frac{1}{2}}$
We plug in the known values to obtain:
$v_f=[(12m/s)^2+2(-35m/s^2)(10.3m)]^{\frac{1}{2}}$
$v_f=8.6m/s$
Thus, we can see that the car's speed is greater than $6.0m/s$.
