Answer
(a) $2.35\frac{m}{s}$
(b) $6.4s$
Work Step by Step
(a) The maximum acceleration of the car can be determined as:
$\Sigma F_x=f_s$
$\implies ma=\mu_smg$
This simplifies to:
$a=\mu_sg$
We plug in the known values to obtain:
$a=(0.24)(9.81)$
$a=2.35\frac{m}{s}$
(b) We know that
$a=\frac{v_f-v_i}{t}$
This can be rearranged as:
$t=\frac{v_f-v_i}{a}$
We plug in the known values to obtain:
$t=\frac{15-0}{2.35}$
$t=6.4s$
