How long does it take the second car to overtake the first car?
Saturday, January 22nd, 2011 at
9:18 pm
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Emily J asked:
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?
Walter
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?
Walter
Filed under: Physics
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Teresa
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..
April
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.
Jimmy
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