Rearranging equations.

[iAmiAdam]

I've left that question, I'll get my teacher to explain it, Thanks for your help anyway, much appreciated.

 

[ColinJ]

lol it's moments like these you wish the forum had laTeX support.

Or that schools had maths teachers!

I've left that question, I'll get my teacher to explain it, Thanks for your help anyway, much appreciated.

Ah, I was beaten to it!

 

[iAmiAdam]

tbf, my gcse maths teacher was very good, maths just doesn't click with me. He stayed behind on days to go through stuff with me but I just couldn't get my head round a lot of it.

 

[ColinJ]

tbf, my gcse maths teacher was very good, maths just doesn't click with me. He stayed behind on days to go through stuff with me but I just couldn't get my head round a lot of it.

I found maths pretty easy, but I struggle with simple tasks that most people find easy! 

Life would be boring if we were all the same... 

 

[TheDoctor]

lol it's moments like these you wish the forum had laTeX support.

No!!!! I don't wish to see the photo of Chuffys' latex support...

 

[marinyork]

Scribbled a few things down for you on what the problem question sounded like iAmiAdam.

Spoiler

Initially it appears a total mess with 10 things. However we know what t[sub]1[/sub] and t[sub]2[/sub] are and v[sub]1[/sub]=u[sub]2[/sub]=v[sub]2[/sub], two others are zero. So we have four unknowns but two can be written in terms of each other straight away. A strategy is to get s[sub]1[/sub], a[sub]1[/sub] and v[sub]1[/sub] (=u[sub]2[/sub]=v[sub]2[/sub]) in terms of each other and then use that we know s[sub]1[/sub] + s[sub]2 [/sub]= 100m
s[sub]1[/sub]=u[sub]1[/sub]t[sub]1[/sub] + (1/2)a[sub]1[/sub]t[sub]1[sup]2[/sup][/sub]
s[sub]1[/sub]=v[sub]1[/sub]t[sub]1[/sub] - (1/2)a[sub]1[/sub]t[sub]1[sup]2[/sup][/sub]
u[sub]1[/sub]t[sub]1[/sub] =0 as u[sub]1[/sub] is 0
v[sub]1[/sub]t[sub]1[/sub] - (1/2)a[sub]1[/sub]t[sub]1[sup]2[/sup] =[/sub][sub][/sub] (1/2)a[sub]1[/sub]t[sub]1[sup]2 [/sup][/sub][sup]
[/sup]v[sub]1[/sub]t[sub]1[/sub] = a[sub]1[/sub]t[sub]1[/sub][sup]2[/sup] ⇔ v[sub]1[/sub] = a[sub]1[/sub]t[sub]1[/sub] so v[sub]1[/sub] =(3/2)a[sub]1
[/sub]
s[sub]1[/sub] = (1/2)* (0+v[sub]1[/sub])*(3/2) ⇔ s[sub]1[/sub]= (3/4) v[sub]1[/sub] but we already know v[sub]1[/sub] =(3/2)a[sub]1[/sub] so s[sub]1[/sub]=(9/8)a[sub]1[/sub] †

Now turn to the second stage s[sub]2[/sub] = 100-s[sub]1[/sub]
u[sub]2[/sub]=v[sub]2[/sub]=v[sub]1[/sub]
v[sub]2[/sub]=u[sub]2[/sub]=v[sub]1[/sub]
a[sub]2[/sub]=0
t[sub]2[/sub]=10.49-1.5 =8.99 seconds

s[sub]2[/sub]= ((u+v)/2)*t[sub]2[/sub]
s[sub]2[/sub]=v[sub]1[/sub]t[sub]2[/sub] as u[sub]2[/sub] + v[sub]2[/sub] =2v[sub]1[/sub] and this cancels with the 2 in the denominator
100-s[sub]1[/sub]=8.99v[sub]1[/sub]
s[sub]1[/sub]=100-8.99v1
but v[sub]1[/sub]=(3/2)a[sub]1[/sub]
s[sub]1[/sub]=100-8.99 * (3/2)a[sub]1[/sub]
from † s[sub]1[/sub] =9/8a[sub]1[/sub]
(9/8) a[sub]1[/sub]= 100-8.99(3/2)a[sub]1[/sub]
100= {(2697/200) + (9/8)}a[sub]1[/sub]
100=(1461/100) a[sub]1[/sub]
a[sub]1[/sub]=10,000/1461 ≈6.845ms[sup]-2[/sup]
From this is follows v[sub]1[/sub] ≈ 10.27ms[sup]-1[/sup] and this sounds about right. You can then find s[sub]1[/sub] and s[sub]2[/sub]

 

[marinyork]

No!!!! I don't wish to see the photo of Chuffys' latex support...

Longers may need cheering up again. It's been a while...

 

[twentysix by twentyfive]

How's it going iAm? MarinY seems to be on the ball.

 

[ColinJ]

Scribbled a few things down for you... [then a spoiler]

That's the first time I've seen those spoiler tags in operation though I've now done a search and found other posts using them.

Are they a custom thing for CycleChat or do they work with most forum software? 

 

[iAmiAdam]

Finished the sheet, handed it in and the teacher said nothing, great stuff, will have to wait till I get it back.

Check this and you'd helped me, I would of done the question had I seen it earlier, but cyclechat is blocked in college.

thanks a lot for that though, must of taken you some time.

 

[iAmiAdam]

I'm back again haha.

Question is; How long for an object, starting from rest, to fall (a) 10m

So far I've got;

S - 10m
u - 0
v - -
a - 9.81 ms[sup]-2[/sup]
t - 1.42s

Worked that out by rearranging s=1/2gt[sup]2 [/sup]to be 2s/g=t[sup]2[/sup]
[sup]
[/sup]
Am I correct to of done that?

 

[ColinJ]

I'm back again haha.

Question is; How long for an object, starting from rest, to fall (a) 10m

So far I've got;

S - 10m
u - 0
v - -
a - 9.81 ms[sup]-2[/sup]
t - 1.42s

Worked that out by rearranging s=1/2gt[sup]2 [/sup]to be 2s/g=t[sup]2[/sup]
[sup]
[/sup]
Am I correct to of done that?

No... but you are correct to have done that!

 

[marinyork]

Worked that out by rearranging s=1/2gt[sup]2 [/sup]to be 2s/g=t[sup]2[/sup]
[sup]
[/sup]
Am I correct to of done that?

Yes. You really need to be more confident in rearranging equations though. Do it stage by stage if necessary until you get used to it. Also draw a diagram and label s=, u= v=, a=, t= and clearly define which direction is positive.

If you're really not confident about rearranging equations I'll write you some out to have a go at. You really need to work at it. Once you're over it you'll then not have any more big problems as you've got a year to work and get used to them.

[sup]


[/sup]

 

[iAmiAdam]

So I'm right yeah? ahaha, Can you tell I didn't take English?


Thanks Marinyork, I got some help last week with it, so I was just making sure I was right before I did the rest of this work. Thanks alot.

 

[ColinJ]

So I'm right yeah? ahaha, Can you tell I didn't take English?

I thought that Maths and English were compulsory subjects, or have the National Curriculum rules changed?