how fast is he going after 3.6 seconds
at = 5.8 * 3.6 = 20.88
so
starting new t = 0 at 20.88 m/s
v = 20.88 + 5.8 t
d = 20.88 t + (5.8/2) t^2
same distance as the other car going 67.4
so
67.4 t = 20.88 t + (5.8/2)t^2
solve quadratic, use reasonable of two answers.
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 5.8 m/s2; 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 67.4 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?
2 answers
67
0 answers