Force acting on a projectile - trying to understand the question

[Femme_physics]

Force acting on a projectile - trying to understand the question...

KABOOM!

I couldn't help it. I just saw a cannon and I had to TRY and solve it from a reason beyond me. Plus, I saw it had calculus in it which I want to try and practice. Sadly, I'm failing to understand parts of the question. For starters, here's the question.

Homework Statement

http://img853.imageshack.us/img853/9849/calculusproblem.jpg

Uploaded with ImageShack.us

The Attempt at a Solution

Well, I don't know what they mean by giving me this equation for the "force acting on the projectile". And they want me to the determine the projectile's velocity when t = t'?

So I need to determine the velocity of the derivative?

 

[cupid.callin]

so if t is time and t at t=0 the gun is just fired ... there is no force? ... that's weird

 

[I like Serena]

This is indeed a calculus question and you're taking quite a jump here.

Still, I can see the attraction, or rather ejection of it.
Let's start with:

Homework Equations

F = m a

v = \int_0^t ~ a ~ dt

s = \int_0^t ~ v ~ dtt' is not a derivative, but some specific time (a constant). It could be for instance t' = 1 second.

And they're asking for the velocity which is the integral of the acceleration.
So what's "a"?

 

[Femme_physics]

This is indeed a calculus question and you're taking quite a jump here.

Still, I can see the attraction, or rather ejection of it.

Are you saying I'm jumping too far? You know what, I have an idea..wait for it

 

[I like Serena]

Are you saying I'm jumping too far? You know what, I have an idea..wait for it

Oh... all right.
Do you have another problem then?
Perhaps one using only derivatives (and not integrals)?

 

[Femme_physics]

Let's stay with this one for now, we'll see how hard it gets, yet! I believe "a" stands for acceleration. I can find it by using

F =ma

Where I plug in for F what they gave me for the force acting on the projectile

http://img121.imageshack.us/img121/8594/acccel.jpg Although I'm not sure what does "C" mean in that force.

 

[I like Serena]

Right! One step down. Only a couple more to go.

C is an as yet unspecified constant, which could be for instance C = 1 [N].
Just as m is an unspecified constant, which could be for instance m = 1 [kg].

Can you integrate this?

 

[Femme_physics]

Woah, this is a lot to take in. Let's do it tomorrow, shall we? I'm curious to know who this will hit! (which we will know by translating the numeral answer to literal!)

 

[I like Serena]

Tomorrow is fine. First thing in the morning?

And I guess I have some bad news.
The problem ends just before the projectile leaves the cannon, so we'll never know what it hit!
Unless we extend the problem...

 

[thepatient]

Aw.. I had already solved for this and wanted to see if I was right. D:

 

[Femme_physics]

Right! One step down. Only a couple more to go.

C is an as yet unspecified constant, which could be for instance C = 1 [N].
Just as m is an unspecified constant, which could be for instance m = 1 [kg].

Can you integrate this?

http://img194.imageshack.us/img194/2860/18587458.jpg

Aw.. I had already solved for this and wanted to see if I was right. D:

Showoff!

 

[mathsTKK]

Hi everyone ! I am a new member in the forum ;) I would like to ask something regarding this question too ^^

Can i find s using the equation
s= ut + 1/2at^2 ?

Thank you everyone ^^

 

[I like Serena]

Let's take this step by step.

Let's start with:
\int ~ \sin(t) ~ dt

You should have a list of standard integrals. Do you?
And is this one on it?

 

[Femme_physics]

Oh, the integral of sin ought to be -cos +0

 

[I like Serena]

Right.

That is, to integrate the expression you would do:
\int_0^t ~ \sin(t) ~ dt = \left[ -\cos(t) \right]_0^t = (-\cos(t)) - (-\cos(0)) = -\cos(t) + 1

Have you seen this before?

Aw.. I had already solved for this and wanted to see if I was right. D:

Patience, we'll get there!

Hi everyone ! I am a new member in the forum ;) I would like to ask something regarding this question too ^^

Can i find s using the equation
s= ut + 1/2at^2 ?

Thank you everyone ^^

Welcome to PF, mathsTKK!

No, you won't be able to use that equation in this problem.
To use that equation you need the speed and the acceleration to be constant.
In this problem they vary, so they are not constant.

 

[Femme_physics]

Have you seen this before?

No...

Getting the feeling I'm getting too ahead of myself and that I need to bolster my calculus before trying out calculus physics.

It's weird, because we did take basic calculus this last semester, but we haven't gotten to apply it, and our mechanics course is over!

Oh well...but yea, getting the feeling I SERIOUSLY need to improve my calculus to solve these sort of stuff.

 

[I like Serena]

Hmmm, didn't you learn something about integrals then?

They're the opposite of taking a derivative.
The only real new thing is that the notation is different (and the notation may seem somewhat daunting at first).

 

[Femme_physics]

Hm, well, I guess the real discouragement is that it's not in my material plan. I'm much more enthused when I know something is in my material plan and I can talk about to other students, and post in my blog Thankfully, this coming Sunday am starting up a fresh new semester so I may have plenty of things to wonder out loud about and ask.

But we learned really basic integrals, we haven't done the chain rule, for instance, and this was the integral I had at the test->http://img808.imageshack.us/img808/4757/thisquestion.jpg

Not too hard, is it?

 

[cupid.callin]

They're the opposite of taking a derivative.
The only real new thing is that the notation is different (and the notation may seem somewhat daunting at first).

And integration is seriously annoying (except for basic questions) ...

 

[I like Serena]

Hm, well, I guess the real discouragement is that it's not in my material plan. I'm much more enthused when I know something is in my material plan and I can talk about to other students, and post in my blog Thankfully, this coming Sunday am starting up a fresh new semester so I may have plenty of things to wonder out loud about and ask.

But we learned really basic integrals, we haven't done the chain rule, for instance, and this was the integral I had at the test->

Not too hard, is it?

Oh, but that's exactly what I wrote down in my last post!
The only difference is that sin(t) was integrated to -cos(t).

But yes, the chain rule is coming up here, which is the last thing you need to be able to solve the problem.

And he, I can understand that you may not want to learn this.

(Although the chain rule is really important in differentiation and integration! )