V entry = a t = 6.3*4.1 = 25.83 m/s
v = 25.83 + 6.3 t
x = 25.83 t + (1/2)(6.3) t^2
for the second car x = 70.7 t
so
70.7 t = 25.83 t + 3.15 t^2
3.15 t^2 - 44.87 t = 0
t (3.15 t -44.87) = 0
so they are at the same x at sthe start of course and at
t = 44.87/3.15
A race driver has made a pit stop to refuel. After refueling, he starts from rest and leaves the pit area with an acceleration whose magnitude is 6.3 m/s2; after 4.1 s he enters the main speedway. At the same instant, another car on the speedway and traveling at a constant velocity of 70.7 m/s overtakes and passes the entering car. The entering car maintains its acceleration. How much time is required for the entering car to catch up with the other car?
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