What is the acceleration of the blue car once the brakes are applied?

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The blue car accelerates uniformly at 4.2 m/s² for 3.2 seconds before maintaining a constant speed of 13.44 m/s for 13.2 seconds. After traveling 198 meters, it applies brakes and comes to a stop over a distance of 22 meters. The acceleration during braking is calculated using the formula Vf² = Vi² + 2ax, but discrepancies in distance measurements may affect the final answer. The discussion highlights the importance of significant figures in online grading systems, as slight variations can lead to incorrect submissions. Ultimately, the correct acceleration of the blue car while braking is approximately -3.8 m/s².
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Homework Statement


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 4.2 m/s2 for 3.2 seconds. It then continues at a constant speed for 13.2 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 220.0 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.

What is the acceleration of the blue car once the brakes are applied

The Attempt at a Solution



I found the distance traveled before the brakes were applied which was 198 m. I also found the speed at which it was constant which came out to be 13.44 m/s. These two values i have been confirmed is correct. I found the distance traveled after the brakes are applied which came out to be 22m (220.0-198).

So then i simply plugged the values into Vf^2=Vi^2+2ax → 0=(13.44)^2+2a(22)and solved for "a" but i got the wrong answer?
 
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Look ok to me.

By the way, what is the yellow car doing in this problem.
 
I also agree that your work looks fine. However, I got 199 m instead of 198 m for the distance to where the blue car applies the brakes. Won't make much difference in the answer for the acceleration, but if you're plugging answers into on online grading system, then it might make a difference if they don't allow much tolerance.

Also, did you include the correct sign for the acceleration in your answer?
 
yes... very picky online grading system indeed. Thanks for your time guys :)
 
TRY this
0=(13)^2+2a(22)
answer is -3.8 or -3.84
the trick with the online grading is usually sig figs and that's why 13 is used not 13.44
 
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