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Mathematics: Post your doubts here!

Binyamine

Binyamine

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Binyamine

Binyamine

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PhyZac said:
View attachment 19870
Assalamu Alikum Wa Rahatullahi Wa Barakatooho.....@Minato112

This is the full question...my problem lies on part (ii) and can i see how the graph looks like of part (iii)

For mark-scheme answer refer the picture below.. And Jazakum Allah Khairan!
View attachment 19871
Click to expand...

(i)

a = -d ; u = 1.4 ; t = 1.2 and v = 1.1

we first find the value of a, the deceleration
v = u + at
a = ( v - u ) / t
= ( 1.1- 1.4 / 1.2
= -0.25

Hence d= 0.25
Distance ; s = (v^2 - u^2)/2 a
= ( (1.1)^2 - (1.4)^2 ) / (2 * -0.25)
= 1.5 m

(ii) Now the ball rebounds and move towards A, and while moving towards A, it stops at C i.e velocity is zero at C

v = 0 ; s = 2 ; a = -0.25

let us first find the initial speed that is the speed with which it rebounds from B towards A

u = square root of ( v^2 - 2 a s )
=square root [ (0)^2 - 2 ( -0.25) ( 2 ) ]
= 1

now we can find the time by using v = u + at

t = (v - u)/ a
= ( 0 - 1 ) / (-0.25)
= 4 seconds.

(iii) I am sure that you have understood how the graph should be. Just to confirm ;
the lines should be straight lines for they are of constant acceleration.

When t=0, velocity was 1.4 hence (0, 1.4) ;
when t=1.2 s, velocity was 1.1 hence (1.2, 1.1) ;

You join these two coordinates.

Now when t = 1.2 ; the ball rebounds at B and starts moving in the opposite direction towards A and comes at rest at C
We calculated the velocity of rebound as being 1 m/s. But since it is moving in the opposite direction; the velocity is infact -1 m/s. recall that velocity is a vector quantity.

So the coordinates (1.2, -1 )
and it comes to rest after a further 4 seconds. Velocity is Zero and the time since the beginning of the experiment is 1.2 + 4 = 5.2

hence coordinates ( 5.2, 0 )
Join these two points by a straight line.
 
P

PhyZac

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Binyamine said:
(i)

a = -d ; u = 1.4 ; t = 1.2 and v = 1.1

we first find the value of a, the deceleration
v = u + at
a = ( v - u ) / t
= ( 1.1- 1.4 / 1.2
= -0.25

Hence d= 0.25
Distance ; s = (v^2 - u^2)/2 a
= ( (1.1)^2 - (1.4)^2 ) / (2 * -0.25)
= 1.5 m

(ii) Now the ball rebounds and move towards A, and while moving towards A, it stops at C i.e velocity is zero at C

v = 0 ; s = 2 ; a = -0.25

let us first find the initial speed that is the speed with which it rebounds from B towards A

u = square root of ( v^2 - 2 a s )
=square root [ (0)^2 - 2 ( -0.25) ( 2 ) ]
= 1

now we can find the time by using v = u + at

t = (v - u)/ a
= ( 0 - 1 ) / (-0.25)
= 4 seconds.

(iii) I am sure that you have understood how the graph should be. Just to confirm ;
the lines should be straight lines for they are of constant acceleration.

When t=0, velocity was 1.4 hence (0, 1.4) ;
when t=1.2 s, velocity was 1.1 hence (1.2, 1.1) ;

You join these two coordinates.

Now when t = 1.2 ; the ball rebounds at B and starts moving in the opposite direction towards A and comes at rest at C
We calculated the velocity of rebound as being 1 m/s. But since it is moving in the opposite direction; the velocity is infact -1 m/s. recall that velocity is a vector quantity.

So the coordinates (1.2, -1 )
and it comes to rest after a further 4 seconds. Velocity is Zero and the time since the beginning of the experiment is 1.2 + 4 = 5.2

hence coordinates ( 5.2, 0 )
Join these two points by a straight line.
Click to expand...
Jazaka Allahu khairan...!!!!! Yes i now understand the question...! And yes i got the look of the graph....!! Thank you so much, Sir. !!!!!! May Allah reward you for the help u gave me and other students Ameen.
 
A

aleezay

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URGENT HELP NEEDED:
this Q isnt directly from Alevel math but it might brush up the skills of all the mathematicians out there. So PLEASE give it a try:

A person is trapped in a dark cave with 3 ways out. One of them leads to safety after 2hrs of travel while the other 2 would return him to the cave after 3 and 5hrs of travel respectively. Assuming that the person is equally likely to opt for any of the openings at all times, what is the expected length of time he spends travelling until reaching safety?
 
A

aleezay

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UrGENT HELP in this one too:

Consider an m * n array of lattice points (points with integer coordinates) on a plane. Determine the no of
a) axis-parallel squares (whose sides are parallel to on of the axis)
b) axis-parallel rectangles
c) squares (not necessarily axis parallel)
d) rectangles (")
e) Triangles
whose vertices are in m*n array and have positive area.
PLEASE be kind enough to solve these before 7a.m tomorrow. Id be highly grateful :)
 
abcde

abcde

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aleezay said:
URGENT HELP NEEDED:
this Q isnt directly from Alevel math but it might brush up the skills of all the mathematicians out there. So PLEASE give it a try:

A person is trapped in a dark cave with 3 ways out. One of them leads to safety after 2hrs of travel while the other 2 would return him to the cave after 3 and 5hrs of travel respectively. Assuming that the person is equally likely to opt for any of the openings at all times, what is the expected length of time he spends travelling until reaching safety?
Click to expand...
Firstly, I apologise for the late reply. :/ You may not need this anymore, but this question is really interesting! :)

Lets say E is the expected time of reaching safety.
=> E = (1/3).2 + (1/3).(3+E) + (1/3).(5+E) (probability of 1st pick + probability of 2nd pick + probability of 3rd pick. E is added in 2nd and 3rd pick cuz the person is in the exact same scenario after travelling those 3 and 5 hours, respectively.)
=> E = [2+ (3+ E) + (5 + E)]/3
=> 3E = 2E + 10
=> E = 10.

So, it would take the person 10 hours (estimated).
 
falcon678

falcon678

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salam!
can anyone help me with how to differentiate and integrate expresion involvin brakets e.g (3x^2+2x)^3
thanks!
 
whitecorp

whitecorp

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falcon678 said:
salam!
can anyone help me with how to differentiate and integrate expresion involvin brakets e.g (3x^2+2x)^3
thanks!
Click to expand...

You would have to employ the chain rule of differentiation.

Generally speaking, if you are asked to differentiate [ f(x) ]^n wrt x,
then the derivative is given by n* [ f(x) ]^(n-1) * f '(x)

If we consider your example (3x^2+2x)^3, then f(x)= 3x^2+2x , n=3 and f '(x)= 6x +2,
which therefore gives us the overall derivative as 3*(3x^2+2x)^2 * (6x+2) (shown)

Hope this helps. Peace.
 
whitecorp

whitecorp

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whitecorp

whitecorp

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Just visiting said:
I need help in how to differentiate sin^3 (X)
The answer is 3sin^2(2X)
Thanks in advance
Click to expand...

Do you mean sin(X) raised to the power of 3? Because if that is the case, the answer you provided is incorrect.

d/dx [ sin^3 (X)]
= d/dx [ sin (X) ] ^3

= 3 [ sin (X) ] ^2 * cos(X) (shown)

For your answer provided, 3sin^2(2X) = 3[ sin(2X) ] ^2
= 3 [ 2 sin(X) cos(X)] ^2

= 12 sin ^2 (X) cos ^2 (X)

This clearly isn't the same as 3 [ sin (X) ] ^2 * cos(X) .

Hope it helps. Peace.
 
sma786

sma786

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Mj 2007. Paper 01..
Question 3.. Prove the identity
...
Help plzz
 

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