WANTED Solutions to SGS Trials 2004 (1 Viewer)

[spoilt brat]

Hey peopleS! i was wondering if anyone has the copy of the solutions to Sydney Grammer School Trials 2004 for 3u maths!... i'm trying to get a hold of it, but i can't seem to find it anywhere!

Thanks heaps!!

 

[withoutaface]

I doubt it.

 

[Rorix]

sorry, I've never heard of this 'Sydney Grammer'.



I've heard of a Sydney Grammar.......


I wonder if they're like sister schools or something?

 

[spoilt brat]

Why do you say that?!??!

 

[Estel]

Rorix holds everyone to high standards

 

[Rorix]

well, i searched on Google for 'Sydney Grammer 3u 2004 trial solutions'

I did my best on the given criteria

 

[nit]

We don't post our trials on the web y'no....

@Rorix: Is this level of grammer good enough?

 

[nit]

Not to spam or anything, but what questions specifically spoilt brat?

 

[spoilt brat]

I'm really sorry for spelling it wrong... ... i didn't mean to. What i meant was obviously 'grammar', sorry. =0( ...

umm.. i was particularly looking at the last question, question 7. Thanks...

 

[Sirius Black]

I gotta stucked on that 2-Q7

Here is the question:
(a)Car A and car B are travelling along a straight line level road at constant speeds Va and Vb repectively. Car A is behind car B but is travelling faster
When car A is exactly D metres behind car B, car A applies its brakes, producing a constant deceleration of k m/s^2
(1) Using calculus, find the speed of car A after it has travelled a distance x metres under braking.
(2) Prove that the cars will collide if Va- Vb >sqrt of (2kD) .

(b)A particle is moving in SHM of period T about a centre O. Its displacement at any time at t is given by x=a*sin(nt), where a is the amplitudes. The point P lies D units on the positive sides of O. Let V be the velocity of the particle when it first passes through P. Show that the time between the first two occasions when the particle passes through P is (T/pi) *inverse tan(VT)/(2PiD).

Heaps of thank you.

 

[spoilt brat]

yeah that's the question!!... any ideas??

 

[mojako]

why am I getting a different answer
(maybe not different, but I just can't get to their expression for V_A - V_B

is part (a) (i)
speed = sqrt(V_A² - 2 k x)
?

Then for part (ii) I put in x = V_B t + D, right??
where t = time after the brake is applied in car A.
and solve for V_A² - 2 k x > 0
??

 

[Sirius Black]

wat's wrong with it?!

why am I getting a different answer
(maybe not different, but I just can't get to their expression for V_A - V_B

is part (a) (i)
speed = sqrt(V_A² - 2 k x)
?

I got this part too
for part (ii) according to the solutn in part (i)
x=V_A²/(2k)
for two cars to collide, then x>V_Bt+D
the critical pt is when V of A is 0 so t (which is the max time taken to collide)=V_A/k
sub all the values in then i got
V_A² > 2V_AV_B+2kD
but it isn't the eqn which is required.
Could anyone tell me that which part did i do incorrectly?

ta

 

[Xayma]

Well from the answer you are missing V_B² on the left hand side, does that help? if not Ill finish it after I get sleep.

 

[Sirius Black]

= =+

Well from the answer you are missing V_B² on the left hand side

I know that i am missing the V_B² bit, but how to get it ?

if not Ill finish it after I get sleep.

would u plz just stay for a little longer and provide some more suggestions?
plz...

 

[Rorix]

Since Sirius asked so nicely ^_^

You should be able to derive
position of A at time t: t*V(a)-kt^2/2 - D (taking A to be at x=-D initally)
position of B: t*V(b)
Distance between A and B:
t(Va-Vb)-kt^2/2-D=0
Rearranging,
kt^2 - 2t(Va-Vb) + 2D = 0
Which is a quadratic in terms of t
We require real and different roots, as if the roots are not different it will just be a instantenous brushing, not a collision
i.e. 4(Va-Vb)^2 > 8kD
Va-Vb>sqrt(2kD)

 

[Sirius Black]

kt^2 - 2t(Va-Vb) + 2D = 0
Which is a quadratic in terms of t
We require real and different roots, as if the roots are not different it will just be a instantenous brushing, not a collision

thank you sooo much Rorix
but y r 2 distinct roots of t required to make the collision take place?

 

[Rorix]

Like, we dont want the bumpers just touching for an instant, we want a HUGE MOTHERFUCKIN 6 DEAD IN 2 EXPLOSIONS FIRE DEPARMENT TOO LATE TO RESPOND BEFORE PROPERTY DAMAGE collision.


Because if there's only 1 root for t, the cars just touch instantenously- there's no collision, like the tangent just touches the circle, it doesn't cut it.

well, i think touching should count as a collision, but the question said > not >= so

 

[Xayma]

Touching instanteously will mean that the rear car does not decellerate due to passings its kinetic energy to the car infront of it.

Ie its likes pulling into a garage and having the bumber just touch the wall, no damage is done. Whereas if you smash into it quickly well you view the windscreen.

 

[Sirius Black]

any ideas on part (b)?

Whew, part a is done (
n1 hav ideas on part b?
I keep getting thingie like (T/2pi) cos^-1 something, rather than tan^-1