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a particle moves in a straight line from a constant point on straight line starting from rest such a=3/8 x^2 find velocity when x = 2

the question is clear and straight forward but when i get the velocity it should be ±√2

what i don't understand is the guide answer refusing the -ve value because acc is always +ve the negative is refused because a > 0

thank you in advance

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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Wasabi
    Jun 4 at 18:00
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No, positive acceleration, alone, does not need to imply positive velocity. Another term for "acceleration in the opposite direction of the velocity" is just "deceleration".

The negative value is being rejected based on physical context. That is, it is being rejected based on the fact you are starting from rest along with the fact that acceleration is always positive.

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    $\begingroup$ As a symbol, "a" can have a negative sign to indicate deceleration, which does not have its own symbol. My question is, does negative velocity exist (what does -30 mph mean)? $\endgroup$
    – r13
    Jun 2 at 3:17
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    $\begingroup$ Velocity is a vector and so is acceleration. So you can define your coordinate axis positive and negative directions to be anything you want, left or right, up or down. The polarity of the velocity and acceleration vectors just follows whatever coordinate system you defined. All negative means is "opposite of the direction you defined as positive". $\endgroup$
    – DKNguyen
    Jun 2 at 3:21
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    $\begingroup$ @r13 You're not being rigorous and are mistaking scalars and magnitudes for vectors. Speed is a scalar, velocity are vectors. Vectors have direction based on an coordinate system (the simplest being vectors in a one dimensional problem like an elevator problem or the OP's problem). Try your approach in an elevator problem, a problem with multiple objects, or with unknowns where you don't know some of the directions ahead of time. Just throw in two particles moving in opposite directions and your approach breaks down mathematically. $\endgroup$
    – DKNguyen
    Jun 2 at 4:15
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    $\begingroup$ Just because the author deemed the problem simple enough to not need to explicitly define a coordinate system doesn't mean it's not there. In English you might say 20m/s to the left and 10ms/ to the right, but if you are defining things using conventional cartesian coordinates the way you put it into the equations is you say -20m/s and +10m/s because "left" and "right" don't mean anything mathematically. It's synonymous. $\endgroup$
    – DKNguyen
    Jun 2 at 4:56
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    $\begingroup$ When the equation spits a number at you, the sign either tells you that your assigned direction for the unknown is opposite of the true direction or it outright tells you the true direction depending on how you went about assigning your unknowns. If choose to assign all unknowns in the positive direction then the polarity of the result gives you the direction. But if you assign unknowns with the direction you think they will be, you are pre-assigning directions and the polarity of the result tells you if your guess was correct or not. Either approach is identical mathematically. $\endgroup$
    – DKNguyen
    Jun 2 at 4:57
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I will rephrase DKNGuyen's answer (which is correct - I upvoted ).

I assume that the positive (+) sign is for position, velocities and acceleration is to the right.

The object is starting from rest. Therefore its velocity is zero. We don't explicitly know the initial position, but we can assume that its a very small value different that zero (any value however small would do and whichever direction $\pm$).

The acceleration is given by:

$$a(x)= \frac{3}{8} x^2 >0 $$

Therefore, because a is greater that zero, the object will accelerate towards the right. There is no position where acceleration can become negative (and therefore at any given time the acceleration can be positive or zero), therefore, the equation:

$$V = V_0 + a\cdot t$$

means that the velocity is always greater than zero (since the initial velocity is $V_0$) in the parameters of this problem.


The other solution (-ve) the post mentions is reserved for the case when:

$$a = \color{red}{\mathbf{-}}\frac 3 8 x^2$$

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    $\begingroup$ You shall review the statement - ".means that the velocity is always greater than zero (since the initial velocity is V0)..." Is V = Vo + at always yield result greater than zero? Not from your comments insisting velocity can be negative, and it has a physical meaning though. $\endgroup$
    – r13
    Jun 2 at 15:21
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    $\begingroup$ Here is a debate on the "terminology" forum you might want to have a look, physics.stackexchange.com/questions/189600/…. $\endgroup$
    – r13
    Jun 2 at 15:31
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    $\begingroup$ Honestly, if we can't agree that positive velocity is how fast an object is moving in one direction, and the physical meaning of negative velocity is how fast an object is moving in the opposite direction, then I am afraid there is not much to debate on an engineering level. $\endgroup$
    – NMech
    Jun 2 at 16:26
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    $\begingroup$ Physics is always up to debate when a physical phenomenon can't be clearly explained. The word "fast" has a positive (+) sense, so do you mean negative velocity means how "slow" an object moves in the same direction, or how "fast" in the opposite direction? $\endgroup$
    – r13
    Jun 2 at 17:07
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    $\begingroup$ I imagine there to be little advances in the fields of science and engineering debating on something as rigorous as the definition of velocity! It’s not hard to grasp, and it is used prevalently in engineering!: if positive velocity means an object is travelling, say, to the right, then of course a larger positive number means the object is going faster to the right. If the velocity is negative, it means the object is travelling to the left. If it is a larger negative number (further from zero), it is going faster to the left. Where do you fail to understand this? $\endgroup$
    – Involute
    Jun 3 at 0:04

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