Two cars start from rest at a red stop light. When the light turns green, both c

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 3.7 seconds. It then continues at a constant speed for 8.1 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 187.3 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.

1) How fast is the blue car going 2.2 seconds after it starts?

2).How fast is the blue car going 8.9 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?

5) What is the total time the blue car is moving?

6) What is the acceleration of the yellow car?

Thanks for the Help!

Explanation / Answer

Given,

ab = 4.6 m/s^2 ; t = 3.7 s ; t1 = 8.1 s ; d = 187.3 m ;

1) t = 2.2 s

v at 2.2 sec accelerating with a = 4.6 m/s2 will be:

v = u + at = 0 + 4.6 x 2.2 = 10.12 m/s

Hence, v = 10.12 m/s

2)t' = 8.9 s

v' = u + a t = 0 + 4.6 x 3.7 = 17.02 m/s

Hence, v' = 17.02 m/s

3) Distance travelled will be:

D = 1/2 a t2 + v' x t1 = 0.5 x 4.6 x 3.7 x 3.7 + 17.02 x 8.1 = 169.35 m

Hence, D = 169.35 m

4) acceleration a' will be:

D' = d - D = 187.3 - 169.35 = 17.95 m

a' = v2/ 2 D' = 17.02 x 17.02 / 2 x 17.95 = 8.07 m/s2

Hence, a' =  8.07 m/s2

5)Time = t2 = v/a' = 17.02/8.07 = 2.11 s

total time = T = t2 + t1 + t = 2.11 + 8.1 + 3.7 = 13.91

Hence, T = 13.91 sec

6)the acc. of the yellow car will be:

ay = 2 d / t2 = 2 x 187.3/13.91 x 13.91 = 1.94 m/s2

Hence, ay =  1.94 m/s2

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.