Quadratic Air Drag and Bullet Motion: Solving for v^2 in Terms of x

[Outrageous]

Homework Statement

A gun is shoot straight up. Assuming that the air drag on the bullet varies quadratically with speed, and the equation of motion for the bullet is (m/2)(dv^2/dx)= -mg-(cv^2).
Find for the downward motion , v^2 in terms of x.
Taking that upward direction is positive.

Homework Equations

The Attempt at a Solution

From the thing I uploaded, is my A statement correct?
is my B, (-V)^2 correct?
in C , that is displacement or distant? I think that is displacement.
Please guide , thanks

 

[Outrageous]

Hope correct orientation

 

[Outrageous]

This?

 

[voko]

You justified the minus before $\frac {dv^2} {dx}$ by saying that the acceleration is downward. You justified the minus before $mg$ in the same way. Thus $- \frac {dv^2} {dx}$ must be negative (to be compatible with the negative $-mg$), thus $\frac {dv^2} {dx}$ must be positive. So the acceleration downward ends up being numerically positive, which is wrong given the sign convention.

When you integrate the equation, I do not understand why you integrate from x to 0 - it should be the other way around.

 

[Outrageous]

You justified the minus before $\frac {dv^2} {dx}$ by saying that the acceleration is downward. You justified the minus before $mg$ in the same way. Thus $- \frac {dv^2} {dx}$ must be negative (to be compatible with the negative $-mg$), thus $\frac {dv^2} {dx}$ must be positive. So the acceleration downward ends up being numerically positive, which is wrong given the sign convention.

When you integrate the equation, I do not understand why you integrate from x to 0 - it should be the other way around.

So do you mean A is correct?
from x to zero because I take upward as positive direction, then the motion of the bullet is downward, is from x to zero, isn't? Do you mean I should take zero to x?

 

[voko]

So do you mean A is correct?

I said "wrong", so it obviously cannot be correct.

from x to zero because I take upward as positive direction, then the motion of the bullet is downward, is from x to zero, isn't? Do you mean I should take zero to x?

When you integrate a differential equation, you integrate both sides from start to end. Where is the start and where is the end here?

 

[Outrageous]

You justified the minus before $\frac {dv^2} {dx}$ by saying that the acceleration is downward. You justified the minus before $mg$ in the same way. Thus $- \frac {dv^2} {dx}$ must be negative (to be compatible with the negative $-mg$), thus $\frac {dv^2} {dx}$ must be positive. So the acceleration downward ends up being numerically positive, which is wrong given the sign .

I am sorry ,not really understand, the acceleration downward ends up positive is exactly the situation when the bullet is downward.
A bit confused ,the acceleration here should take its sign as acceleration or deceleration or up or down ?

 

[Outrageous]

When you integrate a differential equation, you integrate both sides from start to end. Where is the start and where is the end here?

The bullet to upward will start x= 0 , end at x.
The bullet downward starts x ,ends ai x=0 .

 

[voko]

I am sorry ,not really understand, the acceleration downward ends up positive is exactly the situation when the bullet is downward.
A bit confused ,the acceleration here should take its sign as acceleration or deceleration or up or down ?

The sign convention is "positive" for "upward".

The entire term on the left, when evaluated numerically, must be negative because the term on right is always negative during the downward motion.

The minus sign on the left implies that $\frac {dv^2} {dx}$ is positive during the downward motion, which means that the acceleration is numerically positive - which is wrong with the given sign convention.

The bullet to upward will start x= 0 , end at x.
The bullet downward starts x ,ends ai x=0 .

x originally meant "current position". What you do makes it "starting position of the downward motion". If that is what you really mean, I suggest you use some other symbol for that.

 

[voko]

Continuing from the last sentence in #9. What you do also means that $v$ that you find after integration is not the velocity at position x, as defined originally, but the velocity as the shell hits the ground, falling from height x. Is that what you are supposed to find?

 

[Outrageous]

The sign convention is "positive" for "upward".

The entire term on the left, when evaluated numerically, must be negative because the term on right is always negative during the downward motion.

The minus sign on the left implies that $\frac {dv^2} {dx}$ is positive during the downward motion, which means that the acceleration is numerically positive - which is wrong with the given sign convention.

Let's say now I drop the negative on my left , then the answer comes out on the right will be negative ,this tell us that the left term is going downward. If I want to calculate the value of acceleration ,I have to drop the negative sign ( as it only tells direction) , is that correct?

Continuing from the last sentence in #9. What you do also means that $v$ that you find after integration is not the velocity at position x, as defined originally, but the velocity as the shell hits the ground, falling from height x. Is that what you are supposed to find?

Yup.

 

[voko]

Let's say now I drop the negative on my left , then the answer comes out on the right will be negative ,this tell us that the left term is going downward.

The answer you get in this case is that the power in the exponent will be negative. Which makes perfect sense physically.

If I want to calculate the value of acceleration ,I have to drop the negative sign ( as it only tells direction) , is that correct?

I am not sure what negative sign you mean here.

 

[Outrageous]

I am not sure what negative sign you mean here.

I should say in this way, the dv/dt always mean that the accelration, all the signs in the equation only tells the direction. nothing is used to represent deceleration.
Correct?

 

[Outrageous]

Really thank you, voko.
Thanks for explaining to me.

 

[voko]

I should say in this way, the dv/dt always mean that the accelration, all the signs in the equation only tells the direction. nothing is used to represent deceleration.
Correct?

We do not usually separate deceleration from acceleration. It is not even possible in the general case. There are some specific cases when one can talk about pure deceleration, but, in general, we just talk about acceleration. Acceleration is a vector, and the direction of the vector indicates its direction. Signs for direction is a simplification, and this simplification is confusing at times, because you can assign any sign to any direction, and that won't change the physics, but you still have to be consistent.

 

[Outrageous]

in general, we just talk about acceleration. Acceleration is a vector, and the direction of the vector indicates its direction. Signs for direction is a simplification, and this simplification is confusing at times, because you can assign any sign to any direction, and that won't change the physics, but you still have to be consistent.

So the a in my case is only considered as acceleration?
One more to ask, upward is positive.
An object is moving upward, after calculating, a= -5.7 ms^(-2). This means the object accelerates 5.7ms^(-2) in downward direction. The negative doesn't mean deceleration here?

 

[voko]

An object is moving upward, after calculating, a= -5.7 ms^(-2). This means the object accelerates 5.7ms^(-2) in downward direction. The negative doesn't mean deceleration here?

This is one of those cases when you can talk about deceleration. Observe that in this case the direction of acceleration is exactly opposite to the direction of velocity.

 

[HallsofIvy]

So the a in my case is only considered as acceleration?
One more to ask, upward is positive.
An object is moving upward, after calculating, a= -5.7 ms^(-2). This means the object accelerates 5.7ms^(-2) in downward direction. The negative doesn't mean deceleration here?

No. The sign on the acceleration says nothing about whether the motion is upward or downward. That is the sign on the velocity. If the velocity is positive, the object is moving upward. If, in addition, the accelertion is negative, the object is moving upward but more slowly as time passes. If the velocity is negative, the object is moving downward. If, in addition, the acceleration is negative, the object is moving down more rapidly as time passes. "Deceleration" is just acceleration directed opposite to the velocity. Here, as the object is going upward, if acceleration is negative, yes, that is "deceleration". But if the object is going downward, negative acceleration is NOT "deceleration".

(This is all assuming the convention that "positive" is upward, "negative" is downward. While that is most common, it is a "convention". You are free to choose the signs however you want, as long as you are consistent throughout the problem.)

 

[Outrageous]

This is one of those cases when you can talk about deceleration. Observe that in this case the direction of acceleration is exactly opposite to the direction of velocity.

Har? Then in this case ?
Upward as positive, an object is moving downward, a= -4.8 ms^(-2). The object should accelerate downward. How can the negative here indicates decelerating?

 

[voko]

Har? Then in this case ?
Upward as positive, an object is moving downward, a= -4.8 ms^(-2). The object should accelerate downward. How can the negative here indicates decelerating?

Again. It is not the sign, it is the direction. Deceleration is acceleration whose direction is opposite of the direction of velocity.

 

[Outrageous]

Thank you guys ,understand already .acceleration ,a calculated doesn't tell direction.
In positive direction , the negative acceleration means deceleration . In negative direction , the negative acceleration mean acceleration.

 

[voko]

Deceleration is a special kind of acceleration which is directed against velocity. Whether it is negative or positive is completely irrelevant.