When to use negative sign ? [3 equations of motion]

[gaganspidey]

Homework Statement

Example 3.13 :

Example 3.1

Homework Equations

v=u+at
S=ut+1/2at²
v²-u²=2aS

The Attempt at a Solution

I know the solution is in front of me, that's ok but I have a confusion regarding negative & positive signs. In both the examples something is moving vertically upwards & in both cases something is released below to the ground, then why in Example 3.13 initial velocity of the stone is taken as u = +12m/s whereas in Example 3.1 u = -2m/s ?
Also in Example 3.13, displacement = -65 m, why ?
Please clear these doubts of mine in as much detail as possible so that I know which sign to use in which case.

 

[Andrew Mason]

It does not matter which direction you put the + y axis. Usually, one uses up as + and down as -. Just choose a direction as + and the opposite direction will be -.

AM

 

[Doc Al]

I know the solution is in front of me, that's ok but I have a confusion regarding negative & positive signs. In both the examples something is moving vertically upwards & in both cases something is released below to the ground, then why in Example 3.13 initial velocity of the stone is taken as u = +12m/s whereas in Example 3.1 u = -2m/s ?

They just used different sign conventions. In 3.13 they used up as the positive direction; in 3.1 they used down as positive.

Also in Example 3.13, displacement = -65 m, why ?

They are measuring the displacement from the starting point, which they call y = 0. (Where you set the origin of your coordinate system is arbitrary. You could also set the ground level to be y = 0, making your initial position y = 65 m.) So the final displacement is below that, thus -65 m.

Please clear these doubts of mine in as much detail as possible so that I know which sign to use in which case.

I recommend that until you are comfortable with arbitrary sign conventions you stick with a standard one: Let up be positive and down be negative. Why not solve 3.1 using that sign convention and see if you get the right answer?

 

[gaganspidey]

@Doc Al

Oh I see ! Yes I get the same answer by taking upward as positive & downward as negative. Just that in this case we take u=-12m/s, s=65 & g=+9.8m/s²
Then why do they confuse us by using different sign conventions when sticking to a standard one does the job.

Thanks for your help too Andrew Mason !

 

[gaganspidey]

One more thing, they say when a body is released from another body moving upwards, u will be negative [Taking upward as (-ve) & downward as (+ve)] because first it reaches the highest point, i.e, it slightly moves up & then proceeds falling downward. So in this case, we'll take u=(-ve), g =+9.8m/s² & s = (+ve).

Am I right here ?

[hope its not a silly question ]

 

[Doc Al]

One more thing, they say when a body is released from another body moving upwards, u will be negative [Taking upward as (-ve) & downward as (+ve)] because first it reaches the highest point, i.e, it slightly moves up & then proceeds falling downward. So in this case, we'll take u=(-ve), g =+9.8m/s² & s = (+ve).

Am I right here ?

Sounds good to me. The first two--the acceleration and initial velocity--are unambiguous. The sign of the displacement depends on whether it ends up above or below the starting point. If it ends up below (like if it falls to the ground), then s will be positive.

 

[gaganspidey]

^ Got it ! Thanks !