In that case, if initial velocity is 20m/s and g=10m/s², climate the displacement increase in 2 seconds would it is in 40m if acceleration due of gravity is positive upward and the displacement would be 10m if g is negative upwards.So it matters right? i am not acquiring the factor what the convention way to to speak actually..

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edited may 30 "17 at 17:11

Qmechanic♦

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asked may 30 "17 at 16:40

MathejuniorMathejunior

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Suppose we adopt the convention the a street upwards is positive and also a street downwards is negative. Velocity is given by:

$$ v = fracdxdt $$

So if the thing is relocating upwards its place increases, i.e. Gets an ext positive with raising time therefore $dx gt 0$ and also $dt gt 0$. That means an object moving upwards has a **positive** velocity. The same argument tells us that an item moving downwards has $dx lt 0$ and also therefore it has actually a **negative** velocity.

So by picking the sign convention for the street we immediately get a sign convention because that the velocity.

But acceleration is offered by:

$$ a = fracdvdt $$

So now we have a sign convention because that velocity this additionally defines the authorize convention for acceleration. If something is increasing upwards it has $dv gt 0$ and also therefore a optimistic acceleration. Likewise something speeding up downwards has actually a negative acceleration.

In your inquiry you"ve provided the usual convention that ranges up space positive, for this reason the early velocity of $+20$ m/s way the object is relocating upwards. And also since the gravitational acceleration is downwards we need to write $g = -9.81$ m/s$^2$.

All you have to do is feeding these values **with your signs** into your equations of motion and also you will acquire the exactly answer.

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You do often see the gravitational acceleration referred to as simply $9.81$ m/s$^2$ i.e. There is no a sign. This is lazy terminology, and also it means that the **magnitude** the $g$ is $9.81$ m/s$^2$.