Vo = at = 5.5*3.6 = 19.8 m/s.
19.8*t + 0.5*5.5*t^2 = 71.7t,
2.75t^2 + 19.8t - 71.7t = 0,
2.75t^2 - 49.15t = 0.
Use Quadratic Formula and get:
17.9 s.
A race driver has a pit stop to refuel. After refueling, he starts from rest and leaves the pit area with an acceleration whose magnitude is 5.5 m/s^2; after 3.6 s he enters the main speedway. At the same instant, another car on the speedway and traveling at a constant velocity of 71.7 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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3 answers