Elementary kinetics problem - Using calculus is different result from algebra

In summary, In my attempt to solve kinematics using calculus I ran into a problem where the acceleration was not constant. I found a solution using the original kinematics equations.
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  • #2
Hey Fp!

Back to mechanics?

The formulas you're using only apply when acceleration is constant.
See the NOTE at the bottom of the solution.
 
  • #3
v=ds/dt

a=dv/dt=(d^2)s/(dt)^2

are real definitions of acceleration. That is why book has used it. The formula for acceleration you are using comes as special case of this formula. That is special case is when ''acceleration is constant'' and here it is not. So we have to use the 'original' kinematics' equations. Not the kinematics equation used in special case.
 
  • #4
I like Serena said:
Hey Fp!

Back to mechanics?

Hi ILS :)
Yep, just trying to stay sharp...am trying to solve kinematics using calculus now which I never really accomplished before :)
The formulas you're using only apply when acceleration is constant.
See the NOTE at the bottom of the solution.
v=ds/dt

a=dv/dt=(d^2)s/(dt)^2

are real definitions of acceleration. That is why book has used it. The formula for acceleration you are using comes as special case of this formula. That is special case is when ''acceleration is constant'' and here it is not. So we have to use the 'original' kinematics' equations. Not the kinematics equation used in special case.

Why is it not defined in the question then whether acceleration is constant or not?
 
  • #5
Femme_physics said:
Hi ILS :)
Yep, just trying to stay sharp...am trying to solve kinematics using calculus now which I never really accomplished before :)

I know. ;)
So is it course material now?
Femme_physics said:
Why is it not defined in the question then whether acceleration is constant or not?

That is implicit from the formula v=3t2+2t.

If the acceleration were constant, you would have a formula like v=2t.
That is, a formula of the form v=v0+at.
 
  • #6
Femme_physics said:
Why is it not defined in the question then whether acceleration is constant or not?
general equation for acceleration

a=dv/dt

we put a as constant tis gives v=at+c

This means when acceleration is constant v is linearly dependent on 't'. Or we can say that acceleration is constant only when 'v' is linearly dependent on t. In the question it is not. See the question mathematically has given that acceleration is not constant.

"Not everything can be said in words, maths say more''
 
  • #7
I know. ;)
So is it course material now?

Oh., no, for this current degree I'm done with technical mechanics. Except, I am a teacher now of non-calculus mechanics, so I do have to explain to students kinematics without calculus which is just a matter of formulas of course so I'll get them some practice. I figure if I am doing this then I might as well enrich myself.

Also, to get a first degree in engineering I will have to study more advanced mechanics anyway...so I'm adopting it as a hobby :)

That is implicit from the formula v=3t2+2t.

If the acceleration were constant, you would have a formula for like v=2t.
That is, a formula of the form v=v0+at.

Ahh... makes perfect sense now :) thank you.
 
  • #8
I notice that you very carefully labeled all your results in m/s, m, etc, when the problem statement was in ft/s, ft, etc. This will get you in trouble. The US Customary system works every bit as well as SI, despite the disparagement it gets from "scientists." If you live in the US, you really need to learn to use it.
 
  • #9
Femme_physics said:
Oh., no, for this current degree I'm done with technical mechanics. Except, I am a teacher now of non-calculus mechanics, so I do have to explain to students kinematics without calculus which is just a matter of formulas of course so I'll get them some practice. I figure if I am doing this then I might as well enrich myself.

Also, to get a first degree in engineering I will have to study more advanced mechanics anyway...so I'm adopting it as a hobby :)

Good! :approve:

Are you going for a first degree in engineering then?
Or were you going to anyway and am I misunderstanding what you mean with the current degree?
 
  • #10
I notice that you very carefully labeled all your results in m/s, m, etc, when the problem statement was in ft/s, ft, etc. This will get you in trouble. The US Customary system works every bit as well as SI, despite the disparagement it gets from "scientists." If you live in the US, you really need to learn to use it.

Oops, force of habit. You're right.

I live in Israel though, we use SI system.

Are you going for a first degree in engineering then?
Or were you going to anyway and am I misunderstanding what you mean with the current degree?

Oh yea, eventually I'll go for a first degree I just got to finish this practical engineer degree first :) I love these stuff.

"Not everything can be said in words, maths say more''

I like that quote :)

So far my favorite quotes

1) Always trust your visual cues - I Like Serena

2) Just remember, Physics is not botany. It's relationships not classification that drive Physics - sophiecentaur

3) "Not everything can be said in words, maths say more" - DarkXponent
 
  • #11
Femme_physics said:
Oh yea, eventually I'll go for a first degree I just got to finish this practical engineer degree first :) I love these stuff.

Wasn't the practical degree enough for your purposes?
Or do you want more now?
I like that quote :)

So far my favorite quotes

1) Always trust your visual cues - I Like Serena

2) Just remember, Physics is not botany. It's relationships not classification that drive Physics - sophiecentaur

3) "Not everything can be said in words, maths say more" - DarkXponent

Ah, you misquoted me, I said "queues", but I like your version better. >_>
 
Last edited:
  • #12
Ah, you misquoted me, I said "queues", but I like your version better. >_>

:)

Wasn't the practical degree enough for your purposes?
Or do you want more now?

My intellectual pursuit passion conquers me. :) So, no, it's certainly not enough for me!

But we shall see in due time. I just like solving mechanics!
 
  • #13
Femme_physics said:
But we shall see in due time. I just like solving mechanics!

And that 'like' will change into 'love' when you learn calculus.
 
  • #14
Mechanics without calculus is really pretty tedious, although I must admit that I too fell for it as a student. When I got into calculus, it really took off and began to soar for me. I have been doing it for over 50 years now, and I still enjoy it tremendously. It is definitely more fun that most reading for me.
 

1. What is the difference between solving a kinetics problem using calculus and using algebra?

When solving a kinetics problem using algebra, you are using basic mathematical equations and formulas to determine the rate of reaction. This approach assumes that the rate of reaction is constant and does not take into account any changes in the reaction over time. On the other hand, using calculus allows you to consider the changing rate of reaction over time by taking into account factors such as concentration, temperature, and pressure.

2. Why might using calculus lead to a different result in a kinetics problem compared to using algebra?

Using calculus allows for a more accurate and precise determination of the rate of reaction by taking into account changes in the reaction over time. This means that the result may be slightly different compared to using algebra, which assumes a constant rate of reaction.

3. Is one method (calculus or algebra) more accurate than the other in solving kinetics problems?

Neither method is necessarily more accurate than the other. It ultimately depends on the complexity of the kinetics problem and the variables involved. In some cases, using algebra may be sufficient, while in others, calculus may be necessary to obtain a more precise result.

4. Can calculus be used for all types of kinetics problems?

Yes, calculus can be used for all types of kinetics problems, including determining reaction rates, rate laws, and reaction mechanisms. It allows for a more comprehensive and detailed analysis of the reaction, taking into account changes over time.

5. Do I need to have a strong understanding of calculus to solve kinetics problems?

While a basic understanding of calculus is helpful in solving kinetics problems, it is not always necessary. Many simple kinetics problems can be solved using algebra, but for more complex problems, a knowledge of calculus may be necessary to obtain more accurate results.

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