Question 1196072
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A = first car
B = second car


A travels at 200 ft/s
B travels at 180 ft/s


Let car A be the one that stalls since it is the faster car.
If B stalled, then there's no way for B to catch up with A.


In other words, the slower car B goes first followed by the faster car A.


t = number of seconds that car A has driven 
t+40 = number of seconds car B has driven


B has a 40 second head start, so it has driven that many extra seconds compared to car A.


Form the distance equation for car A
distance = rate*time
d = r*t
d = 200t


And do so for car B
d = r*t
d = 180*(t+40)
d = 180t+7200


These distances are in feet.


The two cars will meet up, on the same lap, when they travel the same distance. 
Equate the distance expressions and solve for t.


carA_distance = carB_distance
200t = 180t+7200 
200t-180t = 7200 
20t = 7200 
t = 7200/20
t = 360
t+40 = 360+40 = 400


Car B is on the road for 400 seconds (traveling 180*400 = 72,000 ft) 
Car A is on the road for 360 seconds (traveling 200*360 = 72,000 ft as well)
72,000 feet = (72,000)/(5,280) = 13.636 miles approximately


It takes 360 seconds for the two cars to meet up. After this point, car A overtakes car B.
360 seconds = 360/60 = 6 minutes


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Let's figure out how many feet are in 6 laps.
1 lap = 2.75 mi
6*(1 lap) = 6*(2.75 mi)
6 laps = 16.5 mi


1 mi = 5280 ft
16.5*1 mi = 16.5*5280 ft
16.5 mi = 87,120 ft


Therefore, 
6 laps = 16.5 mi = 87,120 ft
or in short
6 laps = 87,120 ft


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The two cars meet up after traveling 72,000 ft
The cars need to travel 87,120 ft to go 6 full laps.


Since 72,000 < 87,120 is the case, we can see that the cars meet up before the 6 laps are up. 


By the time 6 laps happen, the faster car A will have already pulled ahead of the slower car B. 


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Summary:


Question: How long does it take the first car to overtake the​ second?
Answer: <font color=red>360 seconds</font> (equivalent to 6 minutes)


Question: Which car will be ahead after six laps?
Answer: <font color=red>Car A</font> (the faster car going 200 ft/s)
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