How Does Calculus Explain Bullet Deceleration Due to Air Resistance?

[Dux]

I hope I'm posting this in the right section. What follows is not an actual homework problem, but it is a problem that might be similar to a textbook problem, and it involves calculus that I do not understand. The question is as follows:

Homework Statement

A bullet is fired from a gun. As soon as it leaves the barrel, the bullet begins to decelerate due to air resistance at a rate defined by the following equation:

a = -kv²

The variable "a" is (negative) acceleration.
The variable "v" is instantaneous velocity.
The variable "k" is simply a constant that relates to the particular properties of the bullet.

From this equation, write a function that describes the time it will take the bullet to travel x distance.

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Now, I have asked this question before, and I actually was given a solution by someone at one point that works. It involved integrating the given function twice, but truthfully, I don't understand it at all, and I would really like to. I am hoping that someone on this forum might be able to help me understand the calculus involved with this sort of thing.

The Attempt at a Solution

The solution is this:

t = (1/(V * k)) * (exp(D * k) - 1)

"V" is the initial velocity of the bullet (i.e., the muzzle velocity); "k" is the constant from the original equation; "D" is the distance the bullet traveled; and "exp" is just shorthand for the exponential equation (i.e., e to the power of D * k).

This solution works. I have tested it. But I have no idea how it was obtained, and it's driving me crazy. If anyone can help me understand it, I would be most appreciative!

 

[SteamKing]

How does the velocity of the bullet change due to the air resistance of the bullet? In other words, what is the relationship between velocity, acceleration (or deceleration, in this case), and time?

 

[HallsofIvy]

You are given that a= -kv^2. Of course a, the acceleration is defined as the derivative of velocity with respect to time so this is the differential equation [itex]dv/dt= -kv^2[/itex] which is the same as [itex]dv/v^2= -kdt[/itex]. Integrate both sides of that equation.

 

[Dux]

You are given that a= -kv^2. Of course a, the acceleration is defined as the derivative of velocity with respect to time so this is the differential equation [itex]dv/dt= -kv^2[/itex] which is the same as [itex]dv/v^2= -kdt[/itex]. Integrate both sides of that equation.

And how do you do that? I have zero understanding of calculus.

 

[LCKurtz]

How do you expect to understand something that uses calculus in its derivation if you have "zero understanding" of calculus? Perhaps you should start by taking a calculus course, which is a prerequisite for studying differential equations.