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 3.9 m/s2 for 4.7 seconds. It then continues at a constant speed for 10.2 seconds, before applying the brakes such that the car's speed decreases uniformly coming to rest 254.16 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 3.3 seconds after it starts? 12.87 s Submit Your submissions:12.87 Computed value: 12.87 Feedback: Correct! Submitted: Tuesday, January 15 at 3:22 PM 2) How fast is the blue car going 8.1 seconds after it starts? 18.33 s Submt Your submissions: | 18.33 | ./ Computed value: 18.33 Submitted: Tuesday, January 15 at 3:23 PM Feedback: Correct 3) How far does the blue car travel before its brakes are applied to slow down? 226.4 Submit Your submissions: 226.4 Computed value: 226.4 Feedback: Correct Submitted: Tuesday, January 15 at 3:25 PM

Answer 1

First, calculate everything you can for the blue car.

* distance covered in the first 4.7 seconds:

d1 = 3.9 ∙ 4.7² / 2 = 43.08 m

* velocity at the end of the first 4.7 seconds:

v1 = 3.9 ∙ 4.7 = 18.33 m/s

* distance covered during the second part:

d2 = 18.33 ∙ 10.2 = 187 m

* distance of the third part:

d3 = 254.16 - 187 – 43.08 = 24.11 m

* acceleration in the third part:

0 = 18.33² + 2 ∙ a ∙ 24.11

a = -6.97 m/s²

* calculate the time for the third part:

24.11 = 18.33 ∙ t / 2

t = 2.63 s

* calculate the total time:

T = 4.7 + 10.2 + 2.63 = 17.53 s

4)

that't the accelartion in the third part, so a = -6.97 m/s²

5)

that is T = 17.53 s

6)

254.16 = a ∙ 17.53² / 2

a = 1.65 m/s²