Why Is Deceleration Expressed as a Positive Value in Physics Problems?

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Deceleration is expressed as a positive value in physics to indicate the magnitude of the reduction in speed, even though it is technically a negative acceleration. In the given problem, the car's initial speed is 10 m/s, and it comes to a stop with a calculated acceleration of -0.625 m/s². The positive value of 0.625 m/s² for deceleration reflects the rate at which the car slows down. This convention helps clarify discussions about motion, as negative values can lead to confusion. Thus, while the car experiences negative acceleration, the term deceleration is used positively to denote the magnitude of that change.
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Homework Statement


"A car is approaching traffic lights. The car is traveling with speed 10ms-1. The driver applies the brakes to the car and the car comes to rest with constant deceleration in 16s. Modelling the car as a particle, find the deceleration of the car."


Homework Equations


v = u + at


The Attempt at a Solution


[STRIKE]s =[/STRIKE], u = 10ms-1, v = 0ms-1, a = ?, t = 16s

v = u + at
0 = 10 + 16a
a = -10 ÷ 16
a = -0.625ms-2

The answer given is 0.625ms-2. Why is the answer positive and not negative?
 
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If the particle (the car) is slowing down it would have negative acceleration.

So, to go from acceleration to deceleration multiply by -1.

Therefore the deceleration would be -1 * -0.625 = 0.625

Is that why the answer is positive or have I misunderstood?
 
The car's acceleration is -0.625 units, its deceleration is 0.625 units :) Game of words
You have understood correctly.
 
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As lendav_rott said,
Negative acceleration is simply deceleration and | positive acceleration can be negative deceleration(Can be,but this is rarely used or never used(in my experience))
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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