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Driving home from school one day, you spot a ball rolling out into the street. You brake for 1.20 s, slowing your 950-kg car from 16.0 m/s to 9.50 m/s. (a) What is the average force exerted on your car during breaking? (b) How far did you travel while breaking?

When I solved to get acceleration I got $$\frac{(9.5-16)}{1.2} = -5.42 m/s^2$$

Then to find (a) I did $$\sum{F} = ma$$ $$\sum{F} = (950)(-5.42)$$ $$\sum{F} = -5149 newtons$$

I guess my question is: Can the newtons be negative or are they an absolute value to where there always positive?

Killian McCarthy
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  • See http://physicstasks.eu/382/a-car-and-a-constant-breaking-force – Nosrati Oct 18 '18 at 18:31
  • The magnitude of the force is always positive, but it acts opposite to the direction of motion. Recall that forces are vectors, not scalars. They have magnitude and direction. We take the direction of motion to be positive. Therefore, the force is negative. The sign simply indicates direction. – KM101 Oct 18 '18 at 18:34
  • Force is a vector quantity. It has direction. The magnitude of the force is positive but the direction is opposite from the direction of travel of the car. In a 1D world you can call that direction "negative". – Doug M Oct 18 '18 at 18:36

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