If we throw other upwards v some initial velocity (of course), then it"s the exact same whether the acceleration increase is taken into consideration positive or negative, right?But by convention, the upward direction is considered positive. That"s good, however see-

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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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$^2$.