C.Tully
Badges: 0
Rep:
?
#1
Report Thread starter 5 years ago
#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
reply
username1560589
Badges: 20
Rep:
?
#2
Report 5 years ago
#2
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?
0
reply
1420787
Badges: 3
Rep:
?
#3
Report 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
reply
username1560589
Badges: 20
Rep:
?
#4
Report 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
reply
Phichi
Badges: 11
Rep:
?
#5
Report 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?


Posted from TSR Mobile
0
reply
username1560589
Badges: 20
Rep:
?
#6
Report 5 years ago
#6
(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.
0
reply
Phichi
Badges: 11
Rep:
?
#7
Report 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.
t^2? How would this give you a linear graph?


Posted from TSR Mobile
0
reply
TenOfThem
Badges: 18
Rep:
?
#8
Report 5 years ago
#8
(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
0
reply
username1560589
Badges: 20
Rep:
?
#9
Report 5 years ago
#9
(Original post by Phichi)
t^2? How would this give you a linear graph?


Posted from TSR Mobile
s is linear witht^{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.
0
reply
Phichi
Badges: 11
Rep:
?
#10
Report 5 years ago
#10
(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
0
reply
username1560589
Badges: 20
Rep:
?
#11
Report 5 years ago
#11
(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
0
reply
TenOfThem
Badges: 18
Rep:
?
#12
Report 5 years ago
#12
(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
0
reply
Phichi
Badges: 11
Rep:
?
#13
Report 5 years ago
#13
(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
0
reply
TenOfThem
Badges: 18
Rep:
?
#14
Report 5 years ago
#14
(Original post by Phichi)
Brain dead ignore me.


Posted from TSR Mobile
K
0
reply
Phichi
Badges: 11
Rep:
?
#15
Report 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.


Posted from TSR Mobile
0
reply
username1560589
Badges: 20
Rep:
?
#16
Report 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.


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.
0
reply
1420787
Badges: 3
Rep:
?
#17
Report 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.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Are you travelling in the Uni student travel window (3-9 Dec) to go home for Christmas?

Yes (97)
28.36%
No - I have already returned home (44)
12.87%
No - I plan on travelling outside these dates (65)
19.01%
No - I'm staying at my term time address over Christmas (35)
10.23%
No - I live at home during term anyway (101)
29.53%

Watched Threads

View All