Mathematics: Post your doubts here!

[usama321]

I used v=u+at

how is that not correct....

also can u explain Q5ii of the same paper??

You can not use that. The equations of motion are only applicable when the magnitude of acceleration is constant.

For 5ii) We know that particle Q was thrown 2 seconds later. Also, we will use s = ut + 1/2 a t^2 because it is asking about the time when one particle is 5 m higher than the other. The equation gives us both time and displacement, and is according to the question.

Now, when P is 5m higher than Q, we know that displacement of P - displacement of Q should equal 5. (Get used to this concept as there are a lot of such questions).
Using the equation
s = 17(t+2) - 5(t+2)^2
We use t + 2 because P was thrown two second earlier. Thus, it has been travelling two seconds more than Q. Now the equation for Q would be
s = 7t - 5t^2

Now we will eliminate s as we know Q(s)-P(s) = 5
17(t+2) - 5(t+2)^2 -(7t-5t^2) = 5
Solve this and the value of t would be 0.9s.

Just use v = u + at now. Use t+2 for P and t for Q. You will get the answer

 

[daredevil]

You can not use that. The equations of motion are only applicable when the magnitude of acceleration is constant.

For 5ii) We know that particle Q was thrown 2 seconds later. Also, we will use s = ut + 1/2 a t^2 because it is asking about the time when one particle is 5 m higher than the other. The equation gives us both time and displacement, and is according to the question.

Now, when P is 5m higher than Q, we know that displacement of P - displacement of Q should equal 5. (Get used to this concept as there are a lot of such questions).
Using the equation
s = 17(t+2) - 5(t+2)^2
We use t + 2 because P was thrown two second earlier. Thus, it has been travelling two seconds more than Q. Now the equation for Q would be
s = 7t - 5t^2

Now we will eliminate s as we know Q(s)-P(s) = 5
17(t+2) - 5(t+2)^2 -(7t-5t^2) = 5
Solve this and the value of t would be 0.9s.

Just use v = u + at now. Use t+2 for P and t for Q. You will get the answer

oohh thaankkkuuu!! i'll tag u for more problems

 

[Namehere]

You can not use that. The equations of motion are only applicable when the magnitude of acceleration is constant.

For 5ii) We know that particle Q was thrown 2 seconds later. Also, we will use s = ut + 1/2 a t^2 because it is asking about the time when one particle is 5 m higher than the other. The equation gives us both time and displacement, and is according to the question.

Now, when P is 5m higher than Q, we know that displacement of P - displacement of Q should equal 5. (Get used to this concept as there are a lot of such questions).
Using the equation
s = 17(t+2) - 5(t+2)^2
We use t + 2 because P was thrown two second earlier. Thus, it has been travelling two seconds more than Q. Now the equation for Q would be
s = 7t - 5t^2

Now we will eliminate s as we know Q(s)-P(s) = 5
17(t+2) - 5(t+2)^2 -(7t-5t^2) = 5
Solve this and the value of t would be 0.9s.

Just use v = u + at now. Use t+2 for P and t for Q. You will get the answer

From where did the fives from s = 17(t+2) - 5(t+2)^2 and s = 7t - 5t^2 come from?

 

[Jelleh Belleh]

From where did the fives from s = 17(t+2) - 5(t+2)^2 and s = 7t - 5t^2 come from?

s=ut + (1/2)at^2

 

[Namehere]

s=ut + (1/2)at^2

oh crap, was using 9.81 as g rather than 10... ok ty!

 

[Jelleh Belleh]

oh crap, was using 9.81 as g rather than 10... ok ty!

I've done that many times as well, be careful you don't do that in the exam!

 

[Snowysangel]

I used v=u+at

how is that not correct....

also can u explain Q5ii of the same paper??

In 5(I) why do you multiply the speeds by 2 to get the respective times

 

[daredevil]

In 5(I) why do you multiply the speeds by 2 to get the respective times

because we need the time for going up and then coming down.... which doubles the distance...
if that's wat u're asking

 

[Snowysangel]

because we need the time for going up and then coming down.... which doubles the distance...
if that's wat u're asking

Why do we double the speeds then?

 

[Snowysangel]

because we need the time for going up and then coming down.... which doubles the distance...
if that's wat u're asking

It says in the ms that T= 2(1.7)-2(.7)..why? Those are the speeds not the time

 

[Snowysangel]

because we need the time for going up and then coming down.... which doubles the distance...
if that's wat u're asking

Oh wait they divided the speeds by acceleration! But why did they multiply by 2?

 

[iProcrastinate]

Can't figure this gem out. Any help, please?

 

[salvatore]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_42.pdf
Pleaseee help me with qn no. 7(i). From the marking scheme, why is K.E = 0.5 x Vapproach - Vreturn? Change in K.e is usually (final speed - initial speed), so why have they done the other way round here?
I'll be really grateful for your help
Suchal Riaz or anyone else

 

[Jelleh Belleh]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_43.pdf

Could someone please explain Q6 i) how to find the angle of inclination and Q7 ii) and iii) ?

 

[Lostsoul]

Is it that Maths does not have specimen papers, or am I unable to find?

 

[Suchal Riaz]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_43.pdf

Could someone please explain Q6 i) how to find the angle of inclination and Q7 ii) and iii) ?

 

[Igcse giver]

can u plz tell in 7ii) why did u multiply 0.7 with 450g?

 

[Beyond650]

Can't figure this gem out. Any help, please?

Here's my take on it. Needless to say, please ignore my not-so-good handwriting. Thank you.
http://i.imgur.com/MZy6OPY.jpg
http://i.imgur.com/qTXoo7p.jpg
http://i.imgur.com/fP0pUjT.jpg

 

[Jelleh Belleh]

View attachment 41931

Thanks so much!

 

[daredevil]

plz explain iii and iv