Mathematics: Post your doubts here!

[hizirofaki]

c

It's not too bad as long as you understand that the v/t graph is the graph of the gradient of s/t and remember a few cases.
constant velocity is a straight diagonal line for displacement
straight diagonal line for velocity is increasing gradient graph for displacement
curved graph for velocity is increasing gradient at an increasing/decreasing rate for displacement, depending on the curve in the v/t graph


Because we are doing definite integration. The c cancels out.


There is no 'right' or 'wrong' direction. Positive can be any direction as long as you show motion in the opposite direction as negative.

an u plz sketch the raph for his 7 marks !!!!

 

[farhan141]

It's not too bad as long as you understand that the v/t graph is the graph of the gradient of s/t and remember a few cases.
constant velocity is a straight diagonal line for displacement
straight diagonal line for velocity is increasing gradient graph for displacement
curved graph for velocity is increasing gradient at an increasing/decreasing rate for displacement, depending on the curve in the v/t graph


Because we are doing definite integration. The c cancels out.


There is no 'right' or 'wrong' direction. Positive can be any direction as long as you show motion in the opposite direction as negative.

Can u give me an example to clear my doubt regarding definite integration and how C cancels?

 

[farhan141]

Because we are doing definite integration. The c cancels out.


There is no 'right' or 'wrong' direction. Positive can be any direction as long as you show motion in the opposite direction as negative.

Also solve my question I posted in previous page

 

[DaniyalK]

Can u give me an example to clear my doubt regarding definite integration and how C cancels?

suppose you integrate v and get this
S = 2t^2 - 3t + c
and you know s = 5 when t = 0
so c = 5
S = 2t^2 -3t + 5
now this is the term you will apply limits to so it becomes
Upper limit - lower limit
but whatever limits you use, the 5 will stay 5 because it has nothing to do with t
so whatever the limits are, you will always end up with 5 - 5 = 0 when you do upper limit - lower limit.
Therefore, we can just ignore it.

 

[DaniyalK]

c

an u plz sketch the raph for his 7 marks !!!!

 

[DaniyalK]

Why isn't T2 20s? My values were 10 and 30 but when I inserted value of 21 in second derivative of velocity it was giving me positive so increasing? And when I put 19 it was giving me negative so decreasing?View attachment 53464

Why are you putting them in the second derivative... Just take 1 derivative. That will give you acceleration. Velocity will be increasing when acceleration is positive and decreasing when acceleration is negative.

 

[farhan141]

Why are you putting them in the second derivative... Just take 1 derivative. That will give you acceleration. Velocity will be increasing when acceleration is positive and decreasing when acceleration is negative.

Isn't that u have to find d2y/dx2 to see if it is inc or dec. for that function.


Also, about the 'c'. I don't understand the limits thingy for displacement :3. Isnt it only for volume/area?

 

[DaniyalK]

Isn't that u have to find d2y/dx2 to see if it is inc or dec. for that function.


Also, about the 'c'. I don't understand the limits thingy for displacement :3. Isnt it only for volume/area?

You may not know it but you ARE calculating the area. Area under v/t time graph is the displacement.
Or in that that example, think of it this way, suppose you want to know the displacement between t = 0 and t = 10.
So it will be displacement up till t = 10 - displacement at t = 0
so it will be S(10) - S(0) and the 5s will cancel.
At S(0), only the constant remains as anything containing t becomes 0. That's why you get the correct answer when you plug in just 1 value of t (10 in this case) in S, because the constants cancel. But it's important to understand that what you are actually doing is finding the area under the graph from t = 0 to t = X. But suppose if it asks you to find the displacement from t = 10 to t = 20. Then you subtract the displacement at 20 from the displacement at 10, in other words, you are finding the area under the v/t graph from 10 to 20.

d2y/dx2 tells you the nature of a stationary point. Increasing/decreasing is checked from the gradient, dy/dx.

 

[farhan141]

You may not know it but you ARE calculating the area. Area under v/t time graph is the displacement.
Or in that that example, think of it this way, suppose you want to know the displacement between t = 0 and t = 10.
So it will be displacement up till t = 10 - displacement at t = 0
so it will be S(10) - S(0) and the 5s will cancel.
At S(0), only the constant remains as anything containing t becomes 0. That's why you get the correct answer when you plug in just 1 value of t (10 in this case) in S, because the constants cancel. But it's important to understand that what you are actually doing is finding the area under the graph from t = 0 to t = X. But suppose if it asks you to find the displacement from t = 10 to t = 20. Then you subtract the displacement at 20 from the displacement at 10, in other words, you are finding the area under the v/t graph from 10 to 20.

d2y/dx2 tells you the nature of a stationary point. Increasing/decreasing is checked from the gradient, dy/dx.

You are amazing thanks

Just one more thing. Does the same theory (ignoring c) apply for calculating distance?

 

[DaniyalK]

You are amazing thanks

Just one more thing. Does the same theory (ignoring c) apply for calculating distance?

If you are integrating an equation for speed, yes.

 

[ashcull14]

 

[farhan141]

If you are integrating an equation for speed, yes.

OK thanks.

Another problem

 

[DaniyalK]

View attachment 53473

let p's time = t
then q's time = t - 2
a = 1.75 and u = 0 for both
Displacement of P = 0.875t^2
Displacement of Q = 0.875(t-2)^2
Displacement of P - Displacement of Q = 4.9
0.875t^2 - 0.875(t-2)^2 = 4.9
t = 2.4
but since it's talking about time in terms of Q in the question
t = 2.4 - 2
= 0.4

 

[DaniyalK]

OK thanks.

Another problem
View attachment 53474

The tensions AP and AQ are equal, so the angles on both sides of 3sqrt3 are the same. The angles are (180-90-30)/2 = 30
Now take the direction of the 3sqrt3 force as your plane and resolve forces along it. Look at the horizontal part.
Tcos30 + Tcos30 = 3sqrt3
T = 3

 

[TheJDOG]

View attachment 53473

Here you go
https://www.xtremepapers.com/community/attachments/image-jpg.53420/

 

[farhan141]

View attachment 53473

 

[farhan141]

The tensions AP and AQ are equal, so the angles on both sides of 3sqrt3 are the same. The angles are (180-90-30)/2 = 30
Now take the direction of the 3sqrt3 force as your plane and resolve forces along it. Look at the horizontal part.
Tcos30 + Tcos30 = 3sqrt3
T = 3

Oh missed on the equal angles part, thanks

 

[hizirofaki]

thxx

 

[ashcull14]

let p's time = t
then q's time = t - 2
a = 1.75 and u = 0 for both
Displacement of P = 0.875t^2
Displacement of Q = 0.875(t-2)^2
Displacement of P - Displacement of Q = 4.9
0.875t^2 - 0.875(t-2)^2 = 4.9
t = 2.4
but since it's talking about time in terms of Q in the question
t = 2.4 - 2
= 0.4

thnk u

 

[ashcull14]

View attachment 53475

Here you go
https://www.xtremepapers.com/community/attachments/image-jpg.53420/

THNK U ppl tht ws VERY helpful