# Checking a relation in S and acceleration

1. May 13, 2015

### Raghav Gupta

1. The problem statement, all variables and given/known data

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
2. Relevant equations
dS/dt = v
dv/dt = A

3. 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.

2. May 13, 2015

### 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.

3. May 13, 2015

### Raghav Gupta

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.

4. May 13, 2015

### haruspex

You also have the undifferentiated initial equation.

5. May 14, 2015

### Raghav Gupta

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

6. May 14, 2015

### haruspex

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

7. May 14, 2015

### Raghav Gupta

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

8. May 14, 2015

### haruspex

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

9. May 14, 2015

### ehild

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.

Last edited: May 14, 2015
10. May 14, 2015

### Raghav Gupta

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.

11. May 14, 2015

### ehild

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

12. May 14, 2015

### Raghav Gupta

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.

13. May 14, 2015

### ehild

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

14. May 14, 2015

### Raghav Gupta

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

15. May 14, 2015

### ehild

So the acceleration is inverse
The question is A, in terms of S. Formulate your answer this way.

16. May 14, 2015

### Raghav Gupta

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

17. May 14, 2015

### ehild

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"?

18. May 14, 2015

### Raghav Gupta

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?

19. May 14, 2015

### ehild

It would be nice

20. May 14, 2015

### Raghav Gupta

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.
Also identical has identical or same meaning as same
See