Question
Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates uniformly at a rate of 4.6 m/s2 for 4.4 seconds. It then continues at a constant speed for 8.5 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 247 meters from where it started. The yellow car accelerates uniformly for the entire distance, finally catching the blue car just as the blue car comes to a stop.
2)How fast is the blue car going 10.4 seconds after it starts?
3)How far does the blue car travel before its brakes are applied to slow down?
4)What is the acceleration of the blue car once the brakes are applied?
2)How fast is the blue car going 10.4 seconds after it starts?
3)How far does the blue car travel before its brakes are applied to slow down?
4)What is the acceleration of the blue car once the brakes are applied?
Answers
1. V1 = Vo + at,
V1 = 0 + 4.6*4.4 = 20.24m/s.
2. V2 = V1 = 20.24m/s after 10.4s.
3. D = di + d2,
d1 = Vo*t + 0.5at^2,
d1 = 0 + 0.5*4.6*(4.4)^2 = 44.53m.
d2 = Vt = 20.24 * 8.5 = 172m.
D = 44.53 + 172 = 217m.
4. Ds = 247 - 217 = 30m = Stopping
distance.
a = (Vf^2 - Vo^2) / 2d,
a = (0 -(20.24)^2 / 60 = -6.83m/s^2.
V1 = 0 + 4.6*4.4 = 20.24m/s.
2. V2 = V1 = 20.24m/s after 10.4s.
3. D = di + d2,
d1 = Vo*t + 0.5at^2,
d1 = 0 + 0.5*4.6*(4.4)^2 = 44.53m.
d2 = Vt = 20.24 * 8.5 = 172m.
D = 44.53 + 172 = 217m.
4. Ds = 247 - 217 = 30m = Stopping
distance.
a = (Vf^2 - Vo^2) / 2d,
a = (0 -(20.24)^2 / 60 = -6.83m/s^2.