Velocity with changing direction

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A particle initially moves at 5 m/s and reverses direction to -5 m/s in 1 second, raising questions about the sign convention for velocity. The discussion emphasizes that the initial velocity can be considered either positive or negative, depending on the chosen reference direction, as long as the sign convention is consistent. To solve for displacement, it's necessary to use both the kinematic equation provided and another equation to account for the change in velocity. The confusion arises from differing interpretations of initial and final velocities, highlighting the importance of clarity in defining direction. Consistency in sign convention is crucial for accurate calculations in physics problems involving motion.
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A particle moving at 5m/s reverses its direction in 1s to move at 5m/s in the opposite direction. If its acceleration is constant, what is its displacement from its original position at 1s?

I know how to solve the equation:
1) x = x0 + v0t + ½ at2
2) a is unknown therefore solve for a

My question is how do you know the the initial velocity is negative? I would say the initial velocity is 5m/s and after it changes direction velocity is then -5m/s. The solution says the opposite, i.e. initial velocity is -5m/s and final velocity is 5m/s. How do you know? Is there some hard and fast rule I am unaware of?

Please help.
 
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It doesn't make any difference which direction you choose as plus or minus, as long as you are consistent in your sign convention. Also, you're going to need another equation in addition to (or in substitution of) the one you have shown to get the answer. No matter what, the plus and minus sign will wear down the best of us.
 
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