Kinematics Problem

[Hckyplayer8]

Hello all. I'm stumped on how to set up one of my challenge questions.


I'm thinking this is an integration problem. The definite integral bounded by lower bound 0 and upper bound (?) "insert equation" dt. Am I suppose to just plug in the given v = and integrate that? How am I suppose to know how long the bullet is in the barrel if its not given (that is actually part C) ?

 

[MarkFL]

Yes, we have:

[MATH]v=\d{x}{t}[/MATH]
And so (switching dummy variables and using boundaries as limits):

[MATH]\int_{x_0}^{x(t)} \,dy=\int_{v_0}^{v(t)} v(z)\,dz[/MATH]
Of course to find the acceleration function, we use:

[MATH]a(t)=\d{v}{t}[/MATH]
To determine how long the bullet is in the barrel, you need to equate the acceleration function to zero, since we are told the acceleration of the bullet is zero just as it leaves the barrel.

 

[Hckyplayer8]

Yes, we have:

[MATH]v=\d{x}{t}[/MATH]
And so (switching dummy variables and using boundaries as limits):

[MATH]\int_{x_0}^{x(t)} \,dy=\int_{v_0}^{v(t)} v(z)\,dz[/MATH]
Of course to find the acceleration function, we use:

[MATH]a(t)=\d{v}{t}[/MATH]
To determine how long the bullet is in the barrel, you need to equate the acceleration function to zero, since we are told the acceleration of the bullet is zero just as it leaves the barrel.

Wouldn't acceleration be a quadratic with acceleration being zero when velocity is also zero prior to firing?

If so the length of the barrel is the x component between the zeros?

 

[MarkFL]

The velocity is a quadratic in \(t\), and so the acceleration will be linear in \(t\).

 

[Hckyplayer8]

How does this look so far?

 

[MarkFL]

Let's pick up where I left off, with a correction:

[MATH]\int_{x_0}^{x(t)} \,dy=\int_{t_0}^{t} v(z)\,dz[/MATH]
Let:

[MATH]a=-5.35\times10^7[/MATH]
[MATH]b=2.55\times10^5[/MATH]
We also know:

[MATH]t_0=0[/MATH]
[MATH]x_0=0[/MATH]
[MATH]\int_{0}^{x(t)} \,dy=\int_{0}^{t} az^2+bz\,dz[/MATH]
Now, complete the integrations.

 

[Hckyplayer8]

Let's pick up where I left off, with a correction:

[MATH]\int_{x_0}^{x(t)} \,dy=\int_{t_0}^{t} v(z)\,dz[/MATH]

Running down this post in plain English...

x = position which is related to time so the lower bound has a position zero when time is zero because the bullet has yet to be fired. The upper bound has yet to be determined.

What is dy and what is the dummy variable z representative of?

Let:

[MATH]a=-5.35\times10^7[/MATH]
[MATH]b=2.55\times10^5[/MATH]

When you labeled these a and b I first thought perhaps they were x and y vectors but that doesn't make sense because there is no change in y while the bullet is in the barrel. Then I thought they are the bounds for integration...but that doesn't make sense cause I thought our lower bound is zero.

We also know:

[MATH]t_0=0[/MATH]
[MATH]x_0=0[/MATH]

I understand this. As mentioned previously, when the bullet has yet to be fired, time is zero and position is zero.

[MATH]\int_{0}^{x(t)} \,dy=\int_{0}^{t} az^2+bz\,dz[/MATH]
Now, complete the integrations.

 

[MarkFL]

Running down this post in plain English...

x = position which is related to time so the lower bound has a position zero when time is zero because the bullet has yet to be fired. The upper bound has yet to be determined.

What is dy and what is the dummy variable z representative of?

It is considered bad form to have the limits and integrand/differential in terms of the same variable, and since in a definite integral the variable gets "integrated out" it is considered a "dummy variable" and can be switched. Observe that:

[MATH]\int_a^b f(x)\,dx=\int_a^b f(t)\,dt=F(b)-F(a)[/MATH]
And so I switched \(x\) with \(y\) and \(t\) with \(z\) for the purpose of carrying out the definite integrations.

When you labeled these a and b I first thought perhaps they were x and y vectors but that doesn't make sense because there is no change in y while the bullet is in the barrel. Then I thought they are the bounds for integration...but that doesn't make sense cause I thought our lower bound is zero.

I just used the constants \(a\) and \(b\) to alleviate the need to write out those numbers in scientific notation several times.

 

[Hckyplayer8]

Finding the antiderivative I got (-17833333.33z^3) + (127500z^2) + C.

Writing out the scientific notation helped ensure I treated constants as just that.

What do I set as the upper bound?

 

[MarkFL]

Let's return to where I left off:

[MATH]\int_{0}^{x(t)} \,dy=\int_{0}^{t} az^2+bz\,dz[/MATH]
Integrating we find:

[MATH]x(t)=\frac{a}{3}t^3+\frac{b}{2}t^2[/MATH]
Using definite integrals removes the need to worry with constants of integration. We have our complete function.

Plugging in for \(a\) and \(b\) and rounding as directed, we have:

[MATH]x(t)=-1.78\times10^7t^3+1.28\times10^5t^2[/MATH]
The next step is to find the acceleration function. What do you get?

 

[Hckyplayer8]

Let's return to where I left off:

[MATH]\int_{0}^{x(t)} \,dy=\int_{0}^{t} az^2+bz\,dz[/MATH]
Integrating we find:

[MATH]x(t)=\frac{a}{3}t^3+\frac{b}{2}t^2[/MATH]
Using definite integrals removes the need to worry with constants of integration. We have our complete function.

Plugging in for \(a\) and \(b\) and rounding as directed, we have:

[MATH]x(t)=-1.78\times10^7t^3+1.28\times10^5t^2[/MATH]
The next step is to find the acceleration function. What do you get?

Since I didn't know the upper bound I just treated it as an indefinite integral.

If a(t) = dv/dt ... which is a chain rule problem where dv/dt = dv/dx * dx/dt.

So now I find the derivative of the velocity function?

If that is so, there has to be more cause you don't need the chain rule for the derivative of the given function az^2 + bz

 

[MarkFL]

The acceleration function is simply the time rate of change of the given velocity function:

[MATH]a(t)=\d{v}{t}=2at+b[/MATH]

 

[Hckyplayer8]

I read that as acceration = the derivative of velocity in regard to time = 2

The acceleration function is simply the time rate of change of the given velocity function:

[MATH]a(t)=\d{v}{t}=2at+b[/MATH]

I thought for the Leibiniz Notation, anytime you had a "denominator" that was not in terms of x, it meant the derivative was a composite of two functions?

Regardless, after performing the power rule the acceleration function is -107000000t + 255000.

So the obvious question this greenhorn has is, what is next?

 

[MarkFL]

I read that as acceration = the derivative of velocity in regard to time = 2


I thought for the Leibiniz Notation, anytime you had a "denominator" that was not in terms of x, it meant the derivative was a composite of two functions?

Regardless, after performing the power rule the acceleration function is -107000000t + 255000.

So the obvious question this greenhorn has is, what is next?

Find what \(t\) is when \(a(t)=0\), and this will tell you when the bullet is just leaving the barrel.

 

[Hckyplayer8]

Find what \(t\) is when \(a(t)=0\), and this will tell you when the bullet is just leaving the barrel.

It takes the bullet .00238s to go from an initial velocity of 0 to an acceleration of 0.

Webassign is telling me I'm wrong when I plug in the known a(t).

At this point, I guess I don't even know what the question is asking.

Where the bullet sits prior to being fired is the initial position. At rest the bullet has a velocity of 0 and an acceleration of 0.

Now we have the amount of time it takes the bullet to travel down and exit the barrel, .00238.

 

[MarkFL]

I get:

[MATH]a(t)=(-1.07\times10^8)t+2.55\times10^5[/MATH]
Equating this to zero, I find:

[MATH]t\approx2.38\times10^{-3}[/MATH]
This agrees with your result.

 

[Hckyplayer8]

I get:

[MATH]a(t)=(-1.07\times10^8)t+2.55\times10^5[/MATH]
Equating this to zero, I find:

[MATH]t\approx2.38\times10^{-3}[/MATH]
This agrees with your result.

2.38 x 10^-3 is the answer to part (a) a(t) or is it the answer to (a) position x(t) in the original question?

 

[MarkFL]

I would answer part (a) with:

[MATH]a(t)=(-1.07\times10^8)t+2.55\times10^5[/MATH]
[MATH]x(t)=(-1.78\times10^7)t^3+(1.28\times10^5)t^2[/MATH]

 

[Hckyplayer8]

I would answer part (a) with:

[MATH]a(t)=(-1.07\times10^8)t+2.55\times10^5[/MATH]
[MATH]x(t)=(-1.78\times10^7)t^3+(1.28\times10^5)t^2[/MATH]

I just emailed the professor regarding guidance on why the website won't accept that format.

How does the .00238s help me find the length of the barrel?

 

[MarkFL]

I just emailed the professor regarding guidance on why the website won't accept that format.

How does the .00238s help me find the length of the barrel?

Given that is the time at which the bullet is leaving the barrel, then the position function \(x(t)\), evaluated at that time, will give you the length of the barrel, in meters.