How much time is required for the entering car to catch up with the other car?
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 5.4 m/s2; after 3.8 s he enters the main speedway. At the same instant, another car on the speedway and traveling at a constant velocity of 68.9 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?
August 1st, 2010 at 12:50 am
t = time to catch up
constant speed car distance traveled in (t) = 68.9t
velocity of accelerating car at main speedway entrance = (3.8)(5.4) = 20.5 m/s
accelerating car distance traveled in (t) = Vi(t) + 1/2(5.4)t² = 20.5t + 2.7t²
set distances traveled by the two cars equal
68.9t = 20.5t + 2.7t²
cancel t’s
68.9 = 20.5 + 2.7t
2.7t = 68.9 – 20.5 = 48.4
t = 48.4/2.7 = 17.9 s ANS
November 28th, 2011 at 3:01 pm
I appreciate the way you write and also the theme on your blog. Did you code this by yourself or was it accomplished by a professional? I’m especially relatively impressed.