TEFA: GCE A/L

Feb. 20, 2022, 9:48 p.m.
Winnie Vallonia
Here's the continuation.

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- Feb. 20, 2022, 10:35 p.m.
  Yuyen
  Nice try Winnie
  
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- Feb. 21, 2022, 5:28 p.m.
  Wanyu
  Your solutions are all correct Winnie. That was good
  
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Feb. 20, 2022, 9:45 p.m.
Winnie Vallonia
Good evening. Here's my solution to Maths mechanics June 2018 Q7

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- Feb. 21, 2022, 5:31 p.m.
  Wanyu
  Your solutions are all correct Winnie. But your diagram has a wrong force direction, the tension on the carried should be pointing to the right not to the left.
  
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- Feb. 21, 2022, 5:38 p.m.
  Wanyu
  Your solutions are all correct Winnie. But your diagram has a wrong force direction, the tension on the carriage should be pointing to the right not to the left.
  
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Feb. 20, 2022, 7:39 p.m.
KENG ElSON
Here is the required question to solve (June 2018 Question 7 (work , energy and power)). we await your respective solutions guys

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- Feb. 20, 2022, 8:19 p.m.
  Yuyen
  thanks sir elson
  
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Jan. 29, 2022, 11:29 a.m.
Obam Erna
I have difficulties in integration and differentiation

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Oct. 29, 2021, 11:02 a.m.
Elsonkeng
Finding the solution of $p(x)$ $=4x^4+x^2-3$ and using it to solve the equation $16sin^4x +4sin^2x -12=0$

Solution:

To find the solution of $p(x)$ we do the following.

Let $x^2 =t p(x) =0 \Rightarrow 4t^2 +t -3 =0 \Rightarrow 4t^2 +4t -3t -3 =0 \Rightarrow 4t(t+1)-3(t+1)=0$

$\Rightarrow t = -1, \frac{3}{4}$

But $x^2 = t \Rightarrow x^2 = 3$ $\Rightarrow x = \pm\sqrt\frac{3}{4}$

Solving the eqaution $16sin^4x +4sin^2x -12=0$ , we can deduce that $4(4sin^4x +sin^2x -3)=0$

For $sin^2 x = t$

$\Rightarrow$ the equation becomes that of $p(t)$ , and hence we can deduce the solutions of $sin^2x$ directly

$sin^2x = \frac{3}{4}\Rightarrow sinx = \pm \sqrt3 /2$

$\Rightarrow x = (-1)^n \arcsin (\pm\frac{\sqrt3}{2}) + \pi n$

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Oct. 25, 2021, 7:55 p.m.
Yuyen
Great Work Sir Elson,

well simplified

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April 11, 2021, 5:10 p.m.
Wanyu
continuation

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April 11, 2021, 5:07 p.m.
Wanyu
Assignment

A particle describes motion in a horizontal circle around the inside of a smooth spherical bowl of ra which is fixed with its horizontal. If the reaction of the bowl is 5/2 times the particles weight, what speed does the particle move with.

Find the time taken to go round the circle one time

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