Displacement of a car before it stops

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The discussion focuses on calculating the displacement of a car that decelerates uniformly from a speed of 22.6 m/s to a stop over 6.13355 seconds, with a net force of 8880 N acting against it. The initial attempt at calculating displacement used the equation d = vit + 0.5at^2, resulting in an incorrect value due to a misunderstanding of acceleration's sign. The correct approach recognizes that the acceleration is negative because the car is slowing down. The final calculation should reflect this, leading to the accurate displacement value.
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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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