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hey people,

heres my problem in the synoptic unit of physics there is a graph question which you have to use a equation to find constants from a graph, i dont understand how to apply the y=mx+c to the equation they give i know how to find the gradient and so on but i dont know how to relate y=mx+c

example:

v=krn (n is in subscritpt)

and you plot the graph log of v against log of r , and k and n are constants, i dont understand how to apply the y=mx+c to find the constants i can guess and 9/10 i am rite but i need firm knowlegde, ne help of points to websites would be brilll cause i am really stuck

heres my problem in the synoptic unit of physics there is a graph question which you have to use a equation to find constants from a graph, i dont understand how to apply the y=mx+c to the equation they give i know how to find the gradient and so on but i dont know how to relate y=mx+c

example:

v=krn (n is in subscritpt)

and you plot the graph log of v against log of r , and k and n are constants, i dont understand how to apply the y=mx+c to find the constants i can guess and 9/10 i am rite but i need firm knowlegde, ne help of points to websites would be brilll cause i am really stuck

yes that is what i meant, you have th eright answear but i dont fully flow what you have done is it possible to maybe point me a site or something because i need to get this on lockdown, i know you have to take logs but it doesnt seem that easy, you made it look so simple

I'm seriously worried about this exam!!!

i dont get how to find k and n from y=mx+c i kno that m is the gradient but i dont kno how to use the equation above to find the correct constant, so that i can work out m from any graph given with different variables, i can see you have taken logs but i dont understand how to related y=mc+c to v=kr^n

heres another one

s/v= tb +v/2a

the graph is plotted is v over s/v ( s being distance)

how to does y=mx+c enable you to find m from the equation above?

s/v= tb +v/2a

the graph is plotted is v over s/v ( s being distance)

how to does y=mx+c enable you to find m from the equation above?

hardworker

i dont get how to find k and n from y=mx+c i kno that m is the gradient but i dont kno how to use the equation above to find the correct constant, so that i can work out m from any graph given with different variables, i can see you have taken logs but i dont understand how to related y=mc+c to v=kr^n

Ok. Do you understand the bits I did with the log laws (log AB = log A + log B and log (a^n) = nlog a)? Those are things you've just got to remember (not sure about AQA's syllabus, but we were taught this for Edexcel when dealing with log-log graphs).

After you take logs, you need to use those rules so you have the right-hand side in a form of (constant x variable) + another constant.

nlog r + log k is what you have on the right hand side. r is a variable, so log r is also a variable. k is a constant, so log k is also constant (sorry if that's really obvious, just trying to help).

It might help if you write y = mx + c underneath the log equation:

log v = nlog r + log k

y = mx + c

If you're plotting y against x, whatever's multiplying x is the gradient and the other stuff is the intercept. Same thing with the log equation - you're plotting log v against log r, the constant multiplying log r is the gradient and log k is the intercept.

hardworker

heres another one

s/v= tb +v/2a

the graph is plotted is v over s/v ( s being distance)

how to does y=mx+c enable you to find m from the equation above?

s/v= tb +v/2a

the graph is plotted is v over s/v ( s being distance)

how to does y=mx+c enable you to find m from the equation above?

Plot s/v on the y axis against v on the x axis. The gradient equals 1/2a and the intercept equals tb.

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