hsc practice questions :) --> can u help me wif these?.....and exercise ur brain :) (1 Viewer)

[lillaila]

Ooo, it's long...i hope u can help me as well as urself


1. Determine the ratio in which point C(6,9)divides interval AB, where A is (-1,-5) and B is (3,3)

* I somehow got 2 answers but dunno how to choose which one.
My answers were 7:3 AND 3:7
# ans...........7:3

2. A particle's motion is defined by equation v^2=12+4x-x^2,
where x is displacement from origin in metres and v is velocity in ms^-1. Initially, the particle is 6m to the right of the origin.

a) Show that the particle is moving in SHM.
b) Find the centre, the period and the amplitude of the motion
c) The displacement of the particle at any time t is given by
x+ asin(nt+@)+b
Find the values of @ and b, given 0 </= @ </= 2pi

* I'm totally blind on this question, esp. when the centre of motion isn't the origin.
# ans............a)??
b) respectively 2, 2pi and 4
c) @= pi/2 b= 2

3. Find values of x (0 <_ x <_ pi) such that the series
1 + root3tanx + 3tan^2x + 3root3tan^3 + .... has a limiting sum.

* I know "r" is root3tanx..........and i think -1 < r < 1 for limiting sum.
# ans..........dunno: "5pi/6 < x < pi/6" ?? OR "x= 5pi/6 or pi/6"

4. A population of marsupials has an initial population of 500. Factors which influence population include birthrates, the number of marsupials killed by ferals, the amount of feed and so on.
The change in population, N, is given:
N = 500/ {1 + ke^(-1.5t)} , where k is a constant and t is in months.

a) explain why population will eventually die out.
b) If at t=0, change in population is 1, use formula to find how long it will take for only 100 marsupials to remain. Give ur ans to nearest month.

* a) My tutor says: as t--> infinity, N-->500 so population will stay at 500 and will eventually die out 'cos of starvation, birthrates etc ...cos they mentioned this in quest.
b) i dunno...........i got t=3.5, then later i got t=5
# ans............dunno what did u get?

EASIER QUESTION I THINK:
5. A particle is projected from ground level with an initial velocity of 80ms. It just clears a 2m high wall 25 m from pt of projection. The base of the wall is at same level as pt of projection. g=10.
Calc. the angle(s) of projection to the nearest minute.

Thank you 4 ur time. U've helped me and urself
Laila

 

[freaking_out]

i think more people will look at this if u actually break up ya questions into different threads...

 

[PoLaRbEaR]

1)3m+n =6
-------------
m - n
3m+n = 6m-6n
3m = 7n
m:n = 7:3

2)a)a=d/dxv/2
a = 2-x
a = -1(x-2)
a = -nx
therefore it is SHM

centre : x=2
period : 2pi/n = 2pi seconds
amplitude : v = 12+4x-x = 0
(x-6)(x+2) = 0
x=6, -2
therefore amplitude = 4units

c)

Find the values of @ and b, given 0

wat do you mean given 0??

 

[PoLaRbEaR]

3)*for it to be limiting sum |r|1 [where means less than or equal than]
r = (sqrt3)tanx
|(sqrt3)tanx| 1
*i(sqrt3)tanx 1 OR ii-(sqrt3)tanx 1
itanx 1/(sqrt3)
x = pi/6
ii tanx -1/(sqrt)3
x = 5pi/6
therefore x = pi/6 or 5pi/6

4)a)as t approaches infinity ke^(-1.5t) approaches 0
hence N = 500, which means all dead

b)1=500/[1+k]
k=499
400 = 500/[(1+499e^-1.5t]
-1.5t = ln100 - ln199600
t = 5 months

5derive y = -x(1+tan)/1280 + xtan [where V=80 and g=10]
x = 25, y = 2
2 = -[125-125tan]/256 + 25tan
512 = -125 - 125tan + 6400tan
125tan-6400tan+637=0
let u =tan
u = [64006375.0686]/250
tan = 51.100 or 0.0997
= 8853' or 542'

[think its wrong]

 

[lillaila]

Originally posted by freaking_out
i think more people will look at this if u actually break up ya questions into different threads...

LOLS Okkkkk, i thought of that
Thanks PoLaRbEaR

 

[Skywalker]

LOL no exercising the brain there, tc. That's just copy-paste from the answers....but whatever works I guess.

P.S The answer to question 5 is wrong too.