A car starts from rest and accelerates uniformly at 3.0 m/s2. A second car starts from rest 6.0 seconds later at the same point and accelerates uniformly at 5.0 m/s2. How long does it take the second car to overtake the first car?
CORDELL
Comments
ok..we are not in school ..so whats with the math questions?..I dont know..and it might depend on if the first driver stops at dunkin..
Distance = u*t + (1/2)a*t^2
Distance travelled by both cars is the same. Initial velocity for both cars is 0. Acceleration, a is given for both cars. Assume time for second car is t, so for the first car it is t + 6. Equate the two distances and slove for t.
Assuming the cars are points of 0 length
First car:
s = ut + 1/2 at^2
s = (3/2)t^2
Second car:
s = u(t - 6) + 1/2a(t - 6)^2
s = (5/2)(t - 6)^2
(3/2)t^2 = (5/2)(t - 6)^2
(3/2)t^2 = (5/2)(t^2 - 12t + 36)
t^2 = (5/2)(t^2 - 12t + 36)/(3/2)
t^2 = (5/3)(t^2 - 12t + 36)
t^2 = 5/3t^2 - 20t + 60
2/3t^2 - 20t + 60 = 0
t^2 - 30t + 90 = 0
t^2 - 30t + 90 + 135 - 135 = 0
t^2 - 30t + 225 = 135
(t - 15)^2 = 135
t - 15 = sqrt(135)
t = sqrt(135) + 15
t has to be positive
t = 11.6 + 15
Time = 26.6s from the first car starting. So the second car catches up in 20.6 seconds
