# Linear Relationships

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#1
Hi,
I need to rearrange the equation s=g*t^2/2 into the form y=mx+c where s is the independent variable, t is the dependent variable, and g is a constant.
Need to do for my coursework
thanks
0
5 years ago
#2
Distance travelled from rest under acceleration of g.

If this is for physics, do you also study maths/further maths?
0
5 years ago
#3
S and T don't have a linear relationship. How about exploring the relationship between them taking logarithms? This would allow you to construct a graph and infer the value of G. I'm speculating you're using experimentation to derive a known formula?
0
5 years ago
#4
(Original post by offhegoes)
S and T don't have a linear relationship. How about exploring the relationship between them taking logarithms? This would allow you to construct a graph and infer the value of G. I'm speculating you're using experimentation to derive a known formula?
I would just plot t2 against s, it is simpler.

You could use the log method, but it is overcomplicating the question.
0
5 years ago
#5
(Original post by morgan8002)
I would just plot t2 against s, it is simpler.

You could use the log method, but it is overcomplicating the question.
They don't have a linear relationship like OP asked, logs are not over complicating it.

How did you plan to get the value of g from that?

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0
5 years ago
#6
(Original post by Phichi)
They don't have a linear relationship like OP asked, logs are not over complicating it.

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Plotting t2(not t) against s would give you a linear graph with gradient of g/2 going through the origin.
0
5 years ago
#7
(Original post by morgan8002)
Plotting t2(not t) against s would give you a linear graph with gradient of g/2 going through the origin.
? How would this give you a linear graph?

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0
5 years ago
#8
(Original post by Phichi)
? How would this give you a linear graph?

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Because the function is linear in t^2

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5 years ago
#9
(Original post by Phichi)
? How would this give you a linear graph?

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s is linear with.
as g/2 is constant, you can write
then let

In practice, for each value of t, calculate , then plot x against s.

0
5 years ago
#10
(Original post by TenOfThem)
Because the function is linear in t^2

Still needs an equation of the form y=mx+c

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5 years ago
#11
(Original post by Phichi)
Still needs an equation of the form y=mx+c

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y = s
m = g/2
x = t2
c = 0
0
5 years ago
#12
(Original post by Phichi)
Still needs an equation of the form y=mx+c

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It is

Y is s
X is t^2
M is g/2
C is 0
0
5 years ago
#13
(Original post by TenOfThem)
Because the function is linear in t^2

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0
5 years ago
#14
(Original post by Phichi)

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K
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5 years ago
#15
(Original post by morgan8002)
y = s
m = g/2
x = t2
c = 0
Considering that's the equation already given, and he's been asked to rearrange, it would seem as if perhaps that isn't the answer, that's the point I'm getting at.

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0
5 years ago
#16
(Original post by Phichi)
Considering that's the equation already given, and he's been asked to rearrange, it wouldn't seem as if perhaps that isn't the answer, that's the point I'm getting at.

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If they wanted to they could use the log method and rearrange, but as it comes out so easily through the substitution x = t2, logs are overkill.

It's the simplest answer, so I would go with it.
0
5 years ago
#17
(Original post by morgan8002)
If they wanted to they could use the log method and rearrange, but as it comes out so easily through the substitution x = t2, logs are overkill.

It's the simplest answer, so I would go with it.
On second glance you're right. Logs would overcomplicate matters since we aren't trying to infer an unknown linear relationship, when we already expect to find a quadratic relationship.
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