Car A starts from rest at t=0 and travels along a straight road with a constant acceleration of 6ft/s² until it reachers a speed of 80 ft/s² . Afterwards it maintains this speed. Also,when t=0, car B located 6000 ft down the road is traveling towards A at a constant speed of 60ft/s. Determine the distance traveled by car A when they pass each other.

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  primula answered this

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  Since you know when car A reaches its top speed (rounded to 13s, call it t1), then the distance A travels in this time is:
  
  ?x = v0t1 + 0.5at1²
  = 0 + 0.5(6.0ft/s²)(13s)²
  = 510ft (rounded)
  
  And car B:
  
  ?x = v0t1 + 0.5at1²
  = (60ft/s)(13s) + 0
  = 780ft
  
  Now since they started at 6000ft apart, at 13s they will be separated by a distance of:
  
  6000ft - (510ft + 780ft)
  = 4700ft (rounded)
  
  From the time (call it t2) they are this far apart until they meet, the sum of each of their respective displacements will be 4700ft, or:
  
  ?x(A) + ?x(B) = 4700ft
  v(A)t2 + v(B)t2 = 4700ft
  (80ft/s)t2 + (60ft/s)t2= 4700ft
  (140ft/s)t2 = 4700ft
  t2 = 33s
  
  So, the total displacement of car A is the sum of the displacement it goes through while accelerating uniformly from rest over t1and the displacement it goes through at constant velocity over t2, or:
  
  ?x = 0.5at1² + vt2
  = 0.5(6.0ft/s)(13s)² + (80ft/s)(33s)
  = 3147ft (3100ft rounded)