Checking a relation in S and acceleration

[Raghav Gupta]

Homework Statement

If S² = at²+ 2bt+c, then the acceleration is
A) Directly proportional to S
B) Inversely proportional to S
C) Directly proportional to S²
D) inversely proportional to S³

Homework Equations

dS/dt = v
dv/dt = A

The Attempt at a Solution

Differentiating both sides,
2Sv = 2at + 2b
Then again differentiating,
2SA + 2v² = 2a
So,
S = (a-v²)/A
We can say that B option is correct or not?
as there is also a velocity term which is not constant and can be written as differential of S.

 

[haruspex]

As you suspect, the varying v is a problem. Your two stages of differentiation yield two equations with three variables, A, v and S. Use an equation to eliminate v.

 

[Raghav Gupta]

As you suspect, the varying v is a problem. Your two stages of differentiation yield two equations with three variables, A, v and S. Use an equation to eliminate v.

Which equation should I use?
If I use the first equation which I have differentiated the term t will come which is also not constant.

 

[haruspex]

Which equation should I use?
If I use the first equation which I have differentiated the term t will come which is also not constant.

You also have the undifferentiated initial equation.

 

[Raghav Gupta]

You also have the undifferentiated initial equation.

But from it I am only getting,
$S = \sqrt{at^2+2bt+c}$
What is the use of this?

 

[haruspex]

But from it I am only getting,
$S = \sqrt{at^2+2bt+c}$
What is the use of this?

You can use it to eliminate t, leaving an equation with only accn, S and constants. It may be messy.

 

[Raghav Gupta]

You can use it to eliminate t, leaving an equation with only accn, S and constants. It may be messy.

But the equation is not linear in terms of t, it's quadratic, we will not get a definite value for t.

 

[haruspex]

But the equation is not linear in terms of t, it's quadratic, we will not get a definite value for t.

I think you'll find there's some cancellation. It works out ok, just do it.

 

[ehild]

But the equation is not linear in terms of t, it's quadratic, we will not get a definite value for t.

You can choose the root which can be positive if you are worried about the two roots. And you can do the same with the other root.

 

[Raghav Gupta]

I think you'll find there's some cancellation. It works out ok, just do it.

You can choose the root which can be positive if you are worried about the two roots.

So getting, $$ t = \frac{-b+ \sqrt{b^2-ac+S^2a}}{a} $$
Now what should I do?
Substituting for v will get me a t term which is not constant.

 

[ehild]

You got that v=(at+b)/S Substitute t here, to get v in terms of S.

 

[Raghav Gupta]

You got that v=(at+b)/S Substitute t here, to get v in terms of S.

Hmm okay, there were lot of cancellations
Finally getting
$S = \frac{ac-b^2}{A}$
So S is inversely proportional to A, means v is constant judging from the second differential of my attempt.

 

[ehild]

Hmm okay, there were lot of cancellations
Finally getting
$S = \frac{ac-b^2}{A}$

I do not think it is correct. Check.
And remember, you need the acceleration in terms of S.

 

[Raghav Gupta]

I do not think it is correct. Check.
And remember, you need the acceleration in terms of S.

Oh yes, thanks
I did not wrote the denominator S²
So finally S³ is inversely proportional to A.

 

[ehild]

So the acceleration is inverse

Oh yes, thanks
I did not wrote the denominator S²
So finally S³ is inversely proportional to A.

The question is A, in terms of S. Formulate your answer this way.

 

[Raghav Gupta]

So the acceleration is inverse

The question is A, in terms of S. Formulate your answer this way.

I know that both are same
If acceleration is inversely proportional to S³,
then S³ is also inversely proportional to acceleration.

 

[ehild]

I know that both are same
If acceleration is inversely proportional to S³,
then S³ is also inversely proportional to acceleration.

Yes, but you should answer the question. The two sentences are not the same. And you have to choose among the given answers. Is one of them identical with "S³ is inversely proportional to A"?

 

[Raghav Gupta]

Yes, but you should answer the question. The two sentences are not the same. And you have to choose among the given answers. Is one of them identical with "S³ is inversely proportional to A"?

Yes the 4th or D option is correct.
Are you asking to me for a formula so others learn from this thread of A in terms of S?

 

[ehild]

Yes the 4th or D option is correct.
Are you asking to me for a formula so others learn from this thread of A in terms of S?

It would be nice

 

[Raghav Gupta]

It would be nice

So should I write also how I differentiated in my attempt.
What is product rule of differentiation, what is the meaning of differentiation,
how I got the value of t, how we find quadratic roots etc.?
In general if one wants to learn in this thread he/she should ask questions as we don't know what they are expecting.

Yes, but you should answer the question. The two sentences are not the same. And you have to choose among the given answers. Is one of them identical with "S³ is inversely proportional to A"?

Also identical has identical or same meaning as same
See
https://www.google.co.in/#q=identical+meaning
Would you say 3*1 is not same as 9/3 but they are identical?

 

[ehild]

So should I write also how I differentiated in my attempt.
What is product rule of differentiation, what is the meaning of differentiation,
how I got the value of t, how we find quadratic roots etc.?
In general if one wants to learn in this thread he/she should ask questions as we don't know what they are expecting.

No, you do not need to explain everything but you should show your solution that it can be followed.

Also identical has identical or same meaning as same
See
https://www.google.co.in/#q=identical+meaning
Would you say 3*1 is not same as 9/3 but they are identical?

Yes, I say. If your teacher asks you what is 3*1 would you answer that it is 9/3?

 

[Raghav Gupta]

Yes, I say. If your teacher asks you what is 3*1 would you answer that it is 9/3?

Why not for a joke.

Okay so ultimately the formula$A = \frac{ac- b^2}{S^3}$

Or A ∝ 1/S³

 

[ehild]

Okay so ultimately the formula$A = \frac{ac- b^2}{S^3}$

Or A ∝ 1/S³

It is all right now.

 

[Raghav Gupta]

Thanks ehild by the way. Some anger came in my post 20, which I know is not good.

 

[ehild]

Thanks ehild by the way. Some anger came in my post 20, which I know is not good.

You are welcome. And you could not make me angry

 

[SammyS]

I did this without solving for explicitly solving for t.

Start with your solution in the OP.

Differentiating both sides,
2Sv = 2at + 2b (1)
Then again differentiating,
2SA + 2v² = 2a (2)
So,
S = (a-v²)/A (3)

Solve equation (1) for v which is dS/dt :

$\displaystyle\ v=\frac{at+b}{S}\$​

Rewrite equation (3), solving for acceleration. Use notation d²S/dt² rather than A.

$\displaystyle\ \frac{d^2S}{dt^2}=\frac{a-v^2}{S}\$​

Now substitute the expression for v into this as well as some other manipulation..

$\displaystyle\ \frac{d^2S}{dt^2}=\frac{aS^2}{S^3}-\frac{(at+b)^2}{S^3}\$​

Substitute the original expression for S² into the numerator and do some basic algebra.

$\displaystyle\ \frac{d^2S}{dt^2}=\frac{a(at^2+2bt+c)}{S^3}-\frac{a^2t^2+2abt+b^2}{S^3}\$​

Then there is significant cancellation. No t remains.

 

[ehild]

Very nice, Sammy!

 

[Raghav Gupta]

I did this without solving for explicitly solving for t.

Start with your solution in the OP.

Solve equation (1) for v which is dS/dt :

$\displaystyle\ v=\frac{at+b}{S}\$​

Rewrite equation (3), solving for acceleration. Use notation d²S/dt² rather than A.

$\displaystyle\ \frac{d^2S}{dt^2}=\frac{a-v^2}{S}\$​

Now substitute the expression for v into this as well as some other manipulation..

$\displaystyle\ \frac{d^2S}{dt^2}=\frac{aS^2}{S^3}-\frac{(at+b)^2}{S^3}\$​

Substitute the original expression for S² into the numerator and do some basic algebra.

$\displaystyle\ \frac{d^2S}{dt^2}=\frac{a(at^2+2bt+c)}{S^3}-\frac{a^2t^2+2abt+b^2}{S^3}\$​

Then there is significant cancellation. No t remains.

Thanks Sammy. This was a bit fast way to solve.