B Distance traveled in the nth second (1 Viewer)

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∆sn = sn - s(n-1)
In the distance in nth seconds formula above, I don't understand 'sn' and (n-1) means what? Could you explain it, please?
 

mjc123

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Where did you find this equation? In a textbook? What does the book say? You have been told before, it's your responsibility to tell us what your symbols mean.
Have a think. Distance travelled in the nth second. Is the difference between two quantities. Any idea what those two quantities might be, just from physical intuition?
 
177
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Where did you find this equation? In a textbook? What does the book say? You have been told before, it's your responsibility to tell us what your symbols mean.
Have a think. Distance travelled in the nth second. Is the difference between two quantities. Any idea what those two quantities might be, just from physical intuition?
I am giving the detail information which my book have below
'Consider a particle starting with initial velocity u and moving with uniform acceleration a. Let the displacement of the particle in the nth second of its motion be Snth. Snth = Sn - (sn-1)
When Sn and Sn-1 are displacements are the particle in n and n-1 seconds. The values of Sn and Sn-1 can be obtained by putting t = n and t = n-1 in the equation. S = ut + 1/2 at^2
Snth = u + 1/2 a (2n-1)'
∆sn = sn - s(n-1)
In the distance in nth seconds formula above, I don't understand 'sn' and (n-1) means what? Could you explain it, please?
 

mjc123

Science Advisor
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It's just the same formula using slightly different notation. ∆sn is the same as Snth; sn is the same as Sn and s(n-1) is the same as Sn-1.
 

jbriggs444

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I am giving the detail information which my book have below
'Consider a particle starting with initial velocity u and moving with uniform acceleration a. Let the displacement of the particle in the nth second of its motion be Snth. Snth = Sn - (sn-1)
When Sn and Sn-1 are displacements are the particle in n and n-1 seconds. The values of Sn and Sn-1 can be obtained by putting t = n and t = n-1 in the equation. S = ut + 1/2 at^2
Snth = u + 1/2 a (2n-1)'
∆sn = sn - s(n-1)
In the distance in nth seconds formula above, I don't understand 'sn' and (n-1) means what? Could you explain it, please?
It seems likely that there is some problem with transcription from the textbook (presumably typeset) and ASCII text here on Physics Forums. The resulting notation is clumsy at best.

We have two functions, ##S## and ##S_{nth}##.

The function ##S## is defined as the displacement after a total of n seconds. So ##S(t) = ut + \frac{1}{2}at^2##
The function ##S_{nth}## is defined as the incremental displacement that takes place during the nth second. So ##S_{nth}(t) = S(t+1) - S(t)##

If you write ##S(t+1)## and ##S(t)## as formulas involving u, t and a then you should be able to derive a simple formula for ##S_{nth}(t)##

Note that I disagree with author's decision to use "nth" as part of the function name. The dummy variable in a function has no useful relationship with the function name. It's like writing ##sin_\theta(\theta)##: completely inappropriate. Hence the choice to use something other than n as the dummy variable in this response.
 
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177
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It's just the same formula using slightly different notation. ∆sn is the same as Snth; sn is the same as Sn and s(n-1) is the same as Sn-1.
Yes.
 
177
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It seems likely that there is some problem with transcription from the textbook (presumably typeset) and ASCII text here on Physics Forums. The resulting notation is clumsy at best.

We have two functions, ##S## and ##S_{nth}##.

The function ##S## is defined as the displacement after a total of n seconds. So ##S(t) = ut + \frac{1}{2}at^2##
The function ##S_{nth}## is defined as the incremental displacement that takes place during the nth second. So ##S_{nth}(t) = S(t+1) - S(t)##

If you write ##S(t+1)## and ##S(t)## as formulas involving u, t and a then you should be able to derive a simple formula for ##S_{nth}(t)##

Note that I disagree with author's decision to use "nth" as part of the function name. The dummy variable in a function has no useful relationship with the function name. It's like writing ##sin_\theta(\theta)##: completely inappropriate. Hence the choice to use something other than n as the dummy variable in this response.
Snth(t) = S(t +1) - St
In the formula above given by you, Could you please explain what S(t + 1) and what St means?
 

mjc123

Science Advisor
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I don't get what you don't understand. You quote for your own formula:
"Snth = Sn - (sn-1) When Sn and Sn-1 are displacements are the particle in n and n-1 seconds."
And you can't work it out when you use t instead of n? jbriggs444 even defines S(t) for you. What's your problem?
 

jbriggs444

Science Advisor
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Snth(t) = S(t +1) - St
In the formula above given by you, Could you please explain what S(t + 1) and what St means?
Please use notation consistently. I never wrote "St". Failure to transcribe accurately may explain why the equation supposedly quoted from the textbook is so badly mangled.

The notation where a single letter is followed by a parenthesized expression most commonly denotes function evaluation. The single letter is a function. The parenthesized expression is the function argument. The value denoted is of the function evaluated for the specified argument.

Example: notation for function evaluation

If f(x) = x+1 then f(3) = 4.


Note that the notation is "overloaded" and can have a different meaning depending on context. If the single letter has been declared as a function then the meaning is function evaluation as above. If the single letter has been declared as an ordinary variable then the meaning is taken as multiplication.

Example: multiplication denoted by juxtaposition

if f = 7, a = 2 and b = 3 then f(a+b) = 7(2+3) = 35

In a properly written mathematical work, all functions and variables will be declared before use. As a result, there is no ambiguity.
 
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I don't get what you don't understand. You quote for your own formula:
"Snth = Sn - (sn-1) When Sn and Sn-1 are displacements are the particle in n and n-1 seconds."
And you can't work it out when you use t instead of n? jbriggs444 even defines S(t) for you. What's your problem?
Snth(t) = S(t +1) - St
In the formula above given by you, Could you please explain when to use S(t + 1) and when to use St? and the formula given by me doesn't match the formula ' Snth(t) = S(t +1) - St' [ my formula is ∆sn = sn - s(n-1)]
 

jbriggs444

Science Advisor
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Snth(t) = S(t +1) - St
In the formula above given by you,
Again, I never gave that formula. You misquoted it.
So ##S_{nth}(t) = S(t+1) - S(t)##
If you want to quote that accurately, use the "Reply" action. The entire post (excluding embedded quotes), including LaTeX formatting will appear in your message window. You can then edit it down to just the passage of interest.

If you attempt to quote by highlighting a passage and clicking to select Reply, you will get a poorly formatted result.
 
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177
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It seems likely that there is some problem with transcription from the textbook (presumably typeset) and ASCII text here on Physics Forums. The resulting notation is clumsy at best.
The function ##S## is defined as the displacement after a total of n seconds. So ##S(t) = ut + \frac{1}{2}at^2##
In the formula above, what does S(t) means together? as I know S = displacment and t = time
 

jbriggs444

Science Advisor
6,881
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In the formula above, what does S(t) means together? as I know S = displacment and t = time
The notation -- a symbol followed by a parenthesized expression denotes the value of a function. As I wrote in the passage that you quoted:

"The function ##S## is defined as the displacement after a total of n seconds".

As in all of mathematics, a thing means what is defined to mean. No more, no less. I defined the function S in that passage. Accordingly, it means what the definition says that it means.
 

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