You’re speeding at 85 km/hr when you notice that you’re only 10 m behind the car in front of you,
which is moving at the legal speed limit of 60 km/hr. you slam on your brakes, and your car negatively
accelerates at 4.2 m/s
2
. assuming the other car continues at constant speed, will you collide? if so, at what
relative speed? if not, what will be the distance between the cars at their closest approach?

Your position in meters will, measured relative to the starting point of the car behind you, be 
x1(t) = 10 + 23.61 t - 1/2 4.2 t^2 
his position will be 
x2(t) = 16.67 t 
Hence at any time the separation s(t) will be 
s(t) = x1(t) - x2(t) = 10 + 6.94 t -2.1 t^2 
Now I assume you mean that you will decelerate UNTIl you are driving at the legal speed limit (60 km/h). That will take you: 
16.67 m/s = 23.61m/s - 4.2 m/s^2 * t 
t = 1.65 seconds 
What is the separation at that time? If it is still greater than zero, there will be no collision: 
s(1.65) = 10 + 6.94 *1.65 -2.1 (1.65)^2 = 15.73 meters. 
Hence you will NOT collide. The 1.65 s you calculated was the time needed to brake to the speed of 60 km/h. 

Initial velocity,  u = 4.9 m/sfinal velocity = v m/sinitial height,  h₁ =  245 m    final height, h₂ = 0a = -9.8  m/s²time taken = t secondss = ut + 0.5at²⇒  (h₂-h₁) = ut + 0.5at²⇒  0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get   t = 7.59sv  = u +  at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is  69.48 m/s downward