Why is acceleration positive in this exercise?

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The discussion revolves around calculating the acceleration of a 1400kg car stopping from 35km/h and the force experienced by a 68kg occupant. The calculated acceleration is -2779m/s², indicating deceleration. However, the textbook provides a positive value for the force, suggesting that it is only concerned with the magnitude of the force rather than its direction. This clarification highlights the distinction between negative acceleration due to deceleration and the positive representation of force magnitude. The key takeaway is that the problem focuses on the absolute value of force, which is why a positive answer is expected.
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Homework Statement


Calculate the acceleration of a 1400kg car if it can stop from 35km/h (9.72m/s) on a dime (diameter= 1.7cm) (0.017m). What is the force felt by the 68kg occupant of the car?

Homework Equations


v^2=v0^2+2a(x-x0)
F=ma

The Attempt at a Solution


ok after solving for a in that equation i get -2779m/s^2. when i go to calculate the force felt by the 68kg constituant, i'd just plug everything in, which would mean i'd get a negative force. the back of my book says that the answer is correct but instead of it being negative like i have it it is positive. WHY? why was the acceleration changed?
 
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It's difficult to tell without having the book, but it's likely that it was simply looking for the magnitude of the force, which would be positive.
 
axmls said:
It's difficult to tell without having the book, but it's likely that it was simply looking for the magnitude of the force, which would be positive.
Capture.PNG
the books question
Capture2.PNG


the answer
 
Yes, it looks like they're just looking for the magnitude of the force.
 

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