Calculating Distance Between Stop Signs Using Accelerated Motion Formulas

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A car accelerates from rest at a stop sign, reaching a speed of 29.82 m/s after 7.10 seconds with an acceleration of 4.20 m/s². It then coasts for 2.10 seconds, covering an additional distance of 62.622 meters. The car subsequently decelerates at 3.10 m/s² until it stops at the next sign, requiring calculation of the time and distance for this phase. The total distance between the stop signs is calculated by summing the distances from all three intervals, leading to an estimated total of approximately 304.164 meters. Accurate application of kinematic equations is essential for determining the correct distances in each phase of motion.
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A car starts from rest at a stop sign. It accelerates at 4.20m/s^2 for 7.10s, coasts for 2.10s, and then slows down at a rate of 3.10m/s^2 for the next stop sign. How far apart are the stop signs?

Please help me start this problem. Not sure where to begin. Thanks
 
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1st Interval

Vf= 0 + 4.20m/s^2(7.10s)
= 29.82m/s

29.82m/s x 7.10s
= 211.72m

2nd Interval

29.82m/s x 2.10s
= 62.622m

3rd Interval

Vf = 29.82m/s - 3.10m/s^2(t)
-29.82m/s = -3.10m/s^2(t)

9.62s = t

29.82m/s - 3.10m/s^2(9.62s)
??

Well, do I have the first two intervals correct?

And I am stuck on the third interval... Does it just equal 29.82m, which means the distance between the two stop signs is 211.722+62.622+29.82= 304.164m

Any help would be greatly appreciated.
 
interval 1 is not right...

Remember this is accelerated motion... d = vt doesn't work here unless by v you mean "average velocity"...

what displacement formulas do you have for accelerated motion...

try to apply one here...
 

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