Help! Air Drag

1. Sep 10, 2006

data1217

I got a problem with my notes in introduction to mechanics.
In the 1st picture, why the deceleration should be " a=-g-kv" instead of " a=-g+kv" ?
The air drag should be against the motion. the motion is downward, then the drag should be pointing upward. So I think it should be "positive" instead of "negative"

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2. Sep 11, 2006

HallsofIvy

Staff Emeritus
Acceleration due to drag is always -kv precisely because, as you say, it is always "against the motion". If v is downward, then v itself is negative: -kv, then is positive.

If k= 0.1, v= 4 m/s (upward) then the drag is -kv= -(0.1)(4)= -0.4 N, downward.
If k= 0.1, v= -4 m/s (downward) then the drag is -kv= -(0.1)(-4)= 0.4 N, upward.

3. Sep 11, 2006

data1217

but why the derive says so?
the v is negative, why not the a is not equal to -g+kv in the derive?
is it a more general derive?