DannyN's SM question thread =D

[luffy]

thanks luffy! =D very helpful ^_^

stuck on another...
sin(2cos^-1(2x)) is defined for x an element of (root(2)/4,1/2)
it can be shown that sin(2cos^-1(2x)=ax(square root(1-b^2x^2) where a and b are positive constants
find a and b
sorry about the text i dont know how to write the square root sign

This ones tricky - This would be my solution:

Recall that  

Therefore:


Now, the cos and cos^-1 cancel each other out (for obvious reasons):


Now, I'm not sure if you know how to simplify matters when you have trigonometric functions of an inverse trig function.
In order to understand this, draw a right-angled triangle, and label one side '2x 'and the hypotenuse '1'. Therefore, cos^{-1} of 2x and 1 will produce the adjacent angle. Therefore, sine of that angle will be the third side divided by 1 (i.e. just the third side). So, you must find the value of the third side. (I doubt I explained this well enough - but its best you draw a diagram and see what I mean):

Therefore, the third side, using simple Pythagoras, will be:









Therefore, a = 4 and b = 2.

I apologize in advance if I made any errors or if I didn't explain anything in enough detail. Let me know if I did..

Hope I helped.

[DannyN]

Stuck on another question ,
i can do parts a,b and c, but d confuses me
the answer for d is -------->     N/2 is when the population is increasing most rapidly

i've attached the question below, thanks in advance for the help ^^

[moekamo]

so you should get then find the double derivative and equate to 0 to find when the rate of change of population is maximum, should get:



so solving y''(t)=0 gives , then subbing this into y(t) gives y=N/2

this way is really long and messy so if anyone has a shorter way then please share!

[DannyN]

wow thanks! i forgot about find the double derivative

no the method isnt that long if i use the calc ^^, but do you think this would be in the tech-free exam?

[moekamo]

its possible, but unlikely, i mean its just a really long quotient rule when your differentiating. And solving is quite simple because your solving y''(t)=0.

[DannyN]

hey guys i need help with another question regarding mechanics

The 5.4kg mass is now placed on a rough plane and a driving force 100N acting upon the particle makes an angle of 20 degrees with the horizontal.

a)if the coefficient of friction between the surface and the object is 0.2, find the acceleration of the object correct to one decimal place.

this is what i did -->

100cos20-0.2R=5.4a
R=5.4g-100sin20
100cos20-0.2(5.4g-100sin20)=5.4a
therefore a=16.7m/s^2

can anyone please confirm whether that is correct?

part b
Find the value of angle for which the acceleration of the particle is maximum, correct to one decimal place.
Considering all the other values are still the same.

my working out is:

100cos(x)-0.2[5.4g-100sin(x)]=5.4a
maximum acceleration occurs when 0.2[5.4g-100sin(x)]=0
so i solved and found the angle was approximately 29 degrees, however when i used this angle to find the acceleration it was less than the acceleration in the previous part, so does this mean my approach for the question was wrong? can someone help me please!

also below is an attachment of the diagram

[xZero]

100cos(x)-0.2[5.4g-100sin(x)]=5.4a
maximum acceleration occurs when 0.2[5.4g-100sin(x)]=0

how you know max acceleration occurs when 0.2[5.4g-100sin(x)]=0? try diff the whole expression with respect to x and see if you find a different answer

[DannyN]

sorry, i assumed maximum acceleration occurs when the retarding force is 0 however, i was wrong. and thank you zero! that works!

[DannyN]

Stuck on another question, could you guys help me out please?
A particle is projected up a rough plane inclined at 40 degrees to the horizontal with an initial speed of 16m/s. It comes instantaneously to rest after 2.3 seconds. Give coefficient of friction between the particle and the plane in three decimal places

[luken93]

Firstly, find acceleration.







Resolving forces up the plane;