Two cars are 175m apart, facing each other. At that instant, the first car is traveling at 9.0 km/h [W] and is accelerating at 3.6m/s^2 [W]. The second car is traveling at a constant velocity of 21.5 m/s [E].?

Let the cars hit each other after time #t# from initial state.

The distance #d_1# traveled in the time #t# by the first car moving with initial velocity #u=9\ \text{km/hr}=2.5\ \text{m/s}# & an acceleration #a=3.6\ \text{m/}s^2# is given by second equation of motion,

#d_1=ut+\frac{at^2}{2}#
#d_1=2.5t+\frac{3.6t^2}{2}#
#d_1=2.5t+1.8t^2 \ ......(1)#

The distance #d_2# traveled in the time #t# by the second car moving with constant velocity #v=21.5\ \text{m/s}#

#d_2=v\timest=21.5t\ .........(2)#

At the point of collision, the sum of distances #d_1# & #d_2# traveled by both the cars must be equal to the initial distance #175\ m#

#\therefore d_1+d_2=175#

#2.5t+1.8t^2+21.5t=175#

#1.8t^2+24t-175=0#

#9t^2+120t-875=0#

Solving the above quadratic equation, as follows
#t=\frac{-120\pm\sqrt{120^2-4(9)(-875)}}{2(9)}#
#=\frac{-120\pm30\sqrt{51}}{18}#
But time #t>0# hence, we have

#t=\frac{-120+30\sqrt{51}}{18}=\frac{5\sqrt{51}-20}{3}=5.236\ s#

Hence the cars will hit each other at time #t=5.236\ s#

Now, the speed of first car just before collision at time #t=5.236\ s# is given by first equation of motion,

#v=u+at#

#v=2.5+3.6\cdot 5.236=21.35\ \text{m/s}#