Displacement of a car before it stops

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SUMMARY

The displacement of a 2410 kg car, initially traveling west at 22.6 m/s and subjected to a retarding force of 8880 N to the east, is calculated using the equation d = vit + 0.5at². The car takes 6.13355 seconds to stop, resulting in a negative acceleration of -3.68 m/s². The correct displacement during this time is determined to be 69.396 meters, calculated as d = 22.6 * 6.13355 - 0.5 * 3.68 * (6.13355)².

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Homework Statement



A 2410 kg car traveling to the west at 22.6 m/s slows down uniformly.

It takes 6.13355 seconds for the the car to come to a stop if the force on the car is 8880N to the east.

What is the car's displacement during the time it takes to stop? Answer in m


Homework Equations



d = vit + .5at^2

The Attempt at a Solution



Fnet=m*a=F
=(2410)a=8880
a=3.68m/s2
d=vit-.5at^2
=22.6*6.13355+(.5)(3.68)(6.13355)^2
=207.8387m

its not 207.8387 or 69.396
 
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=22.6*6.13355+(.5)(3.68)(6.13355)^2
Motion is retarding . So acceleration should be negative.
=22.6*6.13355-(.5)(3.68)(6.13355)^2
 

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