Mathematics: Post your doubts here!

[$$AK$$]

V/T2 = acceleration = 0.3
T2=V/0.3

man this was T1!!

 

[$$AK$$]

V/T2 = acceleration = 0.3
T2=V/0.3

thank you
but can u solve it to me (part ii ) THNX

 

[MiniSacBall]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s13_qp_41.pdf

Q3 iii i need help please! Explain.

 

[kitkat <3 :P]

I guess the problem is in the third part. An easy way to do this question is to consider the movement of the signal from its highest point, where its u = 0. Now we know that total time is underroot 17s. Thus when conisdering only the returing time, it would become underroot 17/2s

Now use S = 1/2 (10) (underroot 17/2)square
S = 5 (17/4)'
S = 21.25m

Thus its height above the cliff was 21.25 Now just use the third equation of motion with v = 0 and you will get the answer

MiniSacBall

 

[GCE As and a level]

MiniSacBall

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w13_qp_43.pdf
Q5 plz

 

[GCE As and a level]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_41.pdf

Q3 iii i need help please! Explain.

V/T2 = acceleration = 0.3
T2=V/0.3

nope , that's it for now!

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w13_qp_43.pdf
Q5 plz

 

[Suchal Riaz]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_41.pdf

Q3 iii i need help please! Explain.

the time above the cliff =2* speed when reached at cliff/acceleration
speed at the cliff= 10*√17 / 2 = 5√17
speed at bottom = u =? distance = 40 m, acceleration = g = 10 m/s²
v²=u²+2as
u²=2*10*40-(5√17)²)
u=√1225
=35 m/s

 

[ZaqZainab]

the time above the cliff =2* speed when reached at cliff/acceleration
speed at the cliff= 10*√17 / 2 = 5√17
speed at bottom = u =? distance = 40 m, acceleration = g = 10 m/s²
v²=u²+2as
u²=2*10*40-(5√17)²)
u=√1225
=35 m/s

so much physics like i hope i took mechanics
statistics on the other hand is totally different

 

[$$AK$$]

the time above the cliff =2* speed when reached at cliff/acceleration
speed at the cliff= 10*√17 / 2 = 5√17
speed at bottom = u =? distance = 40 m, acceleration = g = 10 m/s²
v²=u²+2as
u²=2*10*40-(5√17)²)
u=√1225
=35 m/s

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w13_qp_43.pdf
Q5 plz

 

[Thought blocker]

so much physics like i hope i took mechanics
statistics on the other hand is totally different

Still it's tough!

 

[ZaqZainab]

Still it's tough!

its is like all the hard topics of physics made dense in the paper

 

[Thought blocker]

its is like all the hard topics of physics made dense in the paper

 

[$$AK$$]

any other query?

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w13_qp_43.pdf
Q5 plz man
tom is my exam

 

[Suchal Riaz]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w13_qp_43.pdf
Q5 plz

'

 

[Suchal Riaz]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w13_qp_43.pdf
Q5 plz man
tom is my exam

answered on above post.

'

 

[$$AK$$]

answered on above post.

thank you man
thnx alot

 

[naadiaa1]

Who's done mechanics paper 4? want to discuss some problems.

 

[naadiaa1]

who's done with mechanics? want to discuss paper 4

 

[Suchal Riaz]

who's done with mechanics? want to discuss paper 4

there are only three people active on this thread and all are doing M1 paper 4. discuss.

 

[Suchal Riaz]

I am going to sleep now. i hope we all do tomorrows paper better than our expectation.