Year 12s: what will be the first way you research your future options? >>

Linear Relationships

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

Reply 1

username1560589

Is your equation

s = g\frac{t^{2}}{2}

?
Distance travelled from rest under acceleration of g.

If this is for physics, do you also study maths/further maths?

(edited 9 years ago)

Reply 2

1420787

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?

Reply 3

username1560589

Original post by offhegoes

I would just plot t² against s, it is simpler.

You could use the log method, but it is overcomplicating the question.

Reply 4

Phichi

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?

Posted from TSR Mobile

(edited 9 years ago)

Reply 5

username1560589

Original post by Phichi

They don't have a linear relationship like OP asked, logs are not over complicating it.

Posted from TSR Mobile

Plotting t²(not t) against s would give you a linear graph with gradient of g/2 going through the origin.

Reply 6

Phichi

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.

t^2

? How would this give you a linear graph?

Posted from TSR Mobile

Reply 7

TenOfThem

Original post by Phichi

t^2

? How would this give you a linear graph?

Posted from TSR Mobile

Because the function is linear in t^2

It is quadratic in t

Reply 8

username1560589

Original post by Phichi

t^2

? How would this give you a linear graph?

Posted from TSR Mobile

s is linear with

t^{2}

.
as g/2 is constant, you can write

s = mt^{2}

then let

x = t^{2}

\therefore s = mx

In practice, for each value of t, calculate

x = t^{2}

, then plot x against s.

The gradient is g/2.

(edited 9 years ago)

Reply 9

Phichi

Original post by TenOfThem

Because the function is linear in t^2

It is quadratic in t

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

Posted from TSR Mobile

Reply 10

username1560589

Original post by Phichi

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

Posted from TSR Mobile

y = s
m = g/2
x = t²
c = 0

Reply 11

TenOfThem

Original post by Phichi

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

Posted from TSR Mobile

It is

Y is s
X is t^2
M is g/2
C is 0

Reply 12

Phichi

Original post by TenOfThem

Because the function is linear in t^2

It is quadratic in t

Brain dead ignore me.

Posted from TSR Mobile

Reply 13

TenOfThem

Original post by Phichi

Brain dead ignore me.

Posted from TSR Mobile

Reply 14

Phichi

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.

Posted from TSR Mobile

(edited 9 years ago)

Reply 15

username1560589

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.

Posted from TSR Mobile

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.

Reply 16

1420787

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.