Bow and arrows, Newtons second law of physics

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SUMMARY

The discussion centers on the effects of air resistance on the speed of an arrow after it has been shot, specifically in the context of Newton's second law of motion (f = ma). It is established that once the arrow is released, the only force acting on it is air resistance, which causes deceleration. The mathematical representation of air resistance is given by the equation F(air resistance) = -λ * v, leading to the solution v(t) = V0 * e^(-λt), indicating that the arrow's speed decreases over time due to this resistance.

PREREQUISITES
  • Understanding of Newton's second law of motion (f = ma)
  • Basic knowledge of differential equations
  • Familiarity with the concept of air resistance
  • Introductory calculus skills
NEXT STEPS
  • Study the principles of air resistance and drag force in physics
  • Learn about differential equations and their applications in motion
  • Explore the effects of gravity on projectile motion
  • Investigate the mathematical modeling of velocity decay over time
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Students beginning their studies in physics, educators teaching mechanics, and anyone interested in the dynamics of projectile motion and forces acting on moving objects.

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Homework Statement


If I have shot an arrow, would that arrow speed slow down due to air resistance? If so, would that mean that the the acceleration would be de-accelerating?

Homework Equations


f = ma

The Attempt at a Solution


This isn't a a homework problem, this is just something I was wondering about the basic laws

Any who, If I shot an arrow where there is air but no gravity, would the arrow eventually slow down? Because i'd imagine if there was no air than the velocity/speed would increase because the acceleration wouldn't change.

Please elaborate fully, this is my first time taking physics, I just want to learn.
 
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Once you've shot the arrow, you are no-longer applying some force on it, but the air does ! It's only decelerating by the air resistance, assuming you know calculus
F(air resistance) = -λ(some coefficient)*v(velocity) and you end up with a differential equation whose solution is v(t) = V0 * e(-λt) , speed is falling but at least he have one :)
 

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