How Does Changing Sign Conventions Affect the Equations for Motion?

[MathewsMD]

I've attached the question along with its solution. It seems fairly simple yet the answer doesn't seem quite right to me.

It appears as though they let downwards be + and that $m\frac{dv}{dt} = mg + 0.4v$ where v is negative since the ball is on its way up and there is a + instead of a - on the air resistance since the direction of motion is up but the air resistance is opposite (same direction as the positive axis), so then the equation becomes: [1] $m\frac{dv}{dt} = mg - 0.4v_s$ and $v_s$ is simply the speed. This makes sense, but when I switch the case and let upwards be +, then I have: $m\frac{dv}{dt} = -mg - 0.4v$ and if I let $v_s$ = speed, then [2] $m\frac{dv}{dt} = -mg - 0.4v_s$. Given the only difference is a change of assigning a direction as positive, shouldn't [1] and [2] be equivalent? Am I missing something here? It seems as though assigning downwards as + yields a different solution than if upwards is assigned as +...

Any clarifications would be greatly appreciated!

 

[Stephen Tashi]

It appears as though they let downwards be + and that $m\frac{dv}{dt} = mg + 0.4v$

All I see in that image is the handwritten solution beginning -m \frac{dv}{dt} = mg - 0.4 v There is some fragment of red printing that is upside down.

 

[MathewsMD]

All I see in that image is the handwritten solution beginning -m \frac{dv}{dt} = mg - 0.4 v There is some fragment of red printing that is upside down.

Yes, I am just stuck on the derivation of the DE, which is shown in the few lines. The red writing just says: "Note that the velocity is negative when the ball is thrown upwards, and when the velocity is negative, the force of air resistance should be positive (down), hence 0.4v." The rest is just the integrating factor and so on. What the solution says works, but I can't seem to form the analogous DE if I change the coordinate system and let upwards be positive instead. Using the logic provided in the answer, I get a different DE (where both the gravitational force and wind resistance terms have the same signage) and this yields different final results.

 

[Stephen Tashi]

I can't visualize how changing the sign of a constant would change the solution.

Out of curiousity, what convention do you use for gravity. Is g = 9.8 and the force of gravity -mg? Or is g = -9.8 and the force of gravity is +mg ?

 

[MathewsMD]

I can't visualize how changing the sign of a constant would change the solution.

Out of curiousity, what convention do you use for gravity. Is g = 9.8 and the force of gravity -mg? Or is g = -9.8 and the force of gravity is +mg ?

Force of gravity when upwards is positive is -mg, while +mg for when the axis has downwards as positive. The only thing throwing me off is b/c the velocity is upwards, so thus it will be a positive scalar if the positive axis is upwards, and then the equation becomes -mg - 0.4v since v is always positive and the resistance is against the direction of motion so it is negative. But this equation is not the same as the one if the positive axis is downwards. The only reason for a difference is b/c the equation is now +mg + 0.4v, but since v in this case is negative the equation is +mg - 0.4v if taking the absolute value of v, which is what is done in the solution posted oddly.