o cricket 27%I 7:30 PM flipitphysics.com FlipltPhysics Physics 4A Mechanics and
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Question
o cricket 27%I 7:30 PM flipitphysics.com FlipltPhysics Physics 4A Mechanics and Wave Motion Spring 2017 Homework: 1-D Kinematics 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 5.1m/s for 4.3 seconds. It then continues at a constant speed for 12 seconds, before applying the brakes such that the car's speed decreases uniformly coming to rest 364.53 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. How fast is the blue car going 2.6 seconds after it starts? m/s, How fast is the blue car going 11.1 seconds after it starts? How far does the blue car travel before its brakes are applied to slow down? What is the total time the blue car is moving? What is the acceleration of the yellow car?Explanation / Answer
Given,
ab = 5.1 m/s^2 ; t = 4.3 s ; t1 = 12 s ; d = 364.53 m ;
1) t = 2.6 s
v at 2.6 sec accelerating with a = 5.1 m/s2 will be:
v = u + at = 0 + 5.1 x 2.6 = 13.26 m/s
Hence, v = 13.26 m/s
2)t' = 11.1 s
v' = u + a t = 0 + 5.1 x 4.3 = 21.93 m/s
Hence, v' = 21.93 m/s
3) Distance travelled will be:
D = 1/2 a t^2 + v' x t1 = 0.5 x 5.1 x 4.3^2 + 21.93 x 8.1 = 141.45 m
Hence, D = 141.45 m
4) acceleration a' will be:
D' = D - d = 364.53 - 141.45 = 223.08 m
a' = v^2/ 2 D' = 21.93^2/2 x 223.08 = 1.08 m/s2
Hence, a' = 1.08 m/s2
5)Time = t2 = v/a' = 21.93/1.08 = 20.31 s
total time = T = t2 + t1 + t = 20.31 + 12 + 4.3 = 36.61s
Hence, T = 36.61 sec
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