Mathematics: Post your doubts here!

[Saad Mughal]

sindh board

 

[Jelleh Belleh]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_42.pdf

Q7 part 1
Could someone please show the correct sketch of the graph?

 

[usama321]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_42.pdf

Q7 part 1
Could someone please show the correct sketch of the graph?

This should work

 

[Rutzaba]

6. The Girl In The Photograph
One school day, a boy named Tom was sitting in class and doing math. It was six more minutes until after school. As he was doing his homework, something caught his eye.

His desk was next to the window, and he turned and looked to the grass outside. It looked like a picture. When school was over, he ran to the spot where he saw it. He ran fast so that no one else could grab it.

He picked it up and smiled. It had a picture of the most beautiful girl he had ever seen. She had a dress with tights on and red shoes, and her hand was formed into a peace sign.

She was so beautiful he wanted to meet her, so he ran all over the school and asked everyone if they knew her or have ever seen her before. But everyone he asked said “No.” He was devastated.

When he was home, he asked his older sister if she knew the girl, but unfortunately she also said “No.” It was very late, so Tom walked up the stairs, placed the picture on his bedside table and went to sleep.

In the middle of the night Tom was awakened by a tap on his window. It was like a nail tapping. He got scared. After the tapping he heard a giggle. He saw a shadow near his window, so he got out of his bed, walked toward his window, opened it up and followed the giggling. By the time he reached it, it was gone.

The next day again he asked his neighbors if they knew her. Everybody said, “Sorry, no.” When his mother came home he even asked her if she knew her. She said “No.” He went to his room, placed the picture on his desk and fell asleep.

Once again he was awakened by a tapping. He took the picture and followed the giggling. He walked across the road, when suddenly he got hit by a car. He was dead with the picture in his hand.

The driver got out of the car and tried to help him, but it was too late. Suddenly he saw the picture and picked it up.

He saw a cute girl holding up three fingers.

 

[Rutzaba]

Conclusion... dont look hee and there while doing maths xD

 

[Suchal Riaz]

Conclusion... dont look hee and there while doing maths xD

How do u write such thrillers?

 

[mohammad hurani]

please

 

[Suchal Riaz]

please

acceleration=v-u / t; so t=(v-u)/a
time for P to reach maximum height = 17-0 / 10 = 1.7 s
time for P to return to ground = 2*1.7=3.4 s

time for Q to reach maximum height = 7/10=0.7
time for Q to return = 1.4s
the difference in time is adjusted by throwing Q Ts later. so T=3.4-1.4=2s

 

[Exo_luhan]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s13_qp_41.pdf
LAST QUESTION GRAPH PLEASE!!!!!

 

[MiniSacBall]

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

Q2 i.... Isn't the formula supposed to WD by P = Gain in Ke+ Gain in Pe + Work Done against resistance.

Why they didn't use Gain in PE.

 

[Jelleh Belleh]

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

Q2 i.... Isn't the formula supposed to WD by P = Gain in Ke+ Gain in Pe + Work Done against resistance.

Why they didn't use Gain in PE.

There is no change in height, the load is being pulled on a straight horizontal path.

 

[MiniSacBall]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_ms_43.pdf

Q5 (iii)

Am i supposed to draw a neat diagram, everything scales, on a graph or just a kind of rough diagram, with those time and velocity labels without any scale?
Are all the graphs supposed to drawed the same way?

 

[Rutzaba]

How do u write such thrillers?

ideas from googles lol

 

[CЯeScɘnt]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_ms_43.pdf

Q5 (iii)

Am i supposed to draw a neat diagram, everything scales, on a graph or just a kind of rough diagram, with those time and velocity labels without any scale?
Are all the graphs supposed to drawed the same way?

you just have to sktch the diagram with important coordinates indicated with a roughly relationship between them but box and whisker plot and frequency diagrams and histograms should properly indicated.

 

[daredevil]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s13_qp_43.pdf

Q4i

why did it become 0.006??

 

[usama321]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_43.pdf

Q4i

why did it become 0.006??

Integration Integrate acceleration and it becomes velocity. As it starts from rest we know the constant is zero. Equate velocity to 90 and you will get the time

 

[Rutzaba]

acceleration=v-u / t; so t=(v-u)/a
time for P to reach maximum height = 17-0 / 10 = 1.7 s
time for P to return to ground = 2*1.7=3.4 s

time for Q to reach maximum height = 7/10=0.7
time for Q to return = 1.4s
the difference in time is adjusted by throwing Q Ts later. so T=3.4-1.4=2s

Integration Integrate acceleration and it becomes velocity. As it starts from rest we know the constant is zero. Equate velocity to 90 and you will get the time

this guy writes too u know

 

[daredevil]

Integration Integrate acceleration and it becomes velocity. As it starts from rest we know the constant is zero. Equate velocity to 90 and you will get the time

I used v=u+at

how is that not correct....

also can u explain Q5ii of the same paper??

 

[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