CSSA 2003 Question (1 Viewer)

[FLR-IT]

alrite

question 9 b(i) and (ii)

i have the answers from the BOS but i dont understand, it doesnt clarify enough


help help!!


thx

 

[LadyMoon]

what answers did you get?

 

[LadyMoon]

ok:
for the first train the one travelling at 80km/h
(the distance traveelled is the area under the graph)
so treat all the times in hours.
therefore the train travelles for 1/3hours at 80km/h and slows down from 1/4 and comes to stop at 't'.

the distance travelled therefore is : 80x1/3 + (80(t-1/3))1/2
which simplifies to: 26.66667 + 40(t-1/3)

and the next train:
acceleraates to 100km.h over 15 mins. which is 1/4 hours.
and travels at 100km/h for 5 mins (which is 1/12) after that (till time is 20mins).
then it slows down.

therefore the area:
(100x(1/4))1/2 + 100(1/12) + (100(t-1/3))1/2

which simplifies to:
20.83333 + 50(t-1/3).

both are equal. therefore:
20.8333 + 50(t-1/3)= 40(t-1/3) + 26.66667

therefor:
10(t-1/3)=5.833333

therefore:
t-1/3=7/12
therefore
t=11/12 hours.

which is 55mins.

(but i dont agree with the solutons way because they bring in km/h, and minutes. It is not a corrct way of doing it)

hope that cleared it up.

 

[FLR-IT]

Originally posted by LadyMoon

the distance travelled therefore is : 80x1/3 + (80(t-1/3))1/2
which simplifies to: 26.66667 + 40(t-1/3)

(100x(1/4))1/2 + 100(1/12) + (100(t-1/3))1/2

wat is with multiplying by the halves ?????

 

[FLR-IT]

OHH I GET IT (ahaha note the time it took me to get it)

thx LADY moon, really appreciate ur help

 

[LadyMoon]

Yeah..... areas of triangle

 

[Wahlito]

Can someone explain question 4(f) to me?

 

[Supra]

i cant remeba wot 4f was, i dont have my paper on hand..
Can u type it out(or main details)

 

[LadyMoon]

4(f): Population of a coastal town is increasing at a decreasing rate.

comment of dp/dt, and d2p/dt2.

Answer:
if you picture the graph, it will look like a half upside down parabola.
(refer to the picture)
therefore as the graph increases the gradient goes from a large steep to a small steep.
and intially the gradient is positive, and it is increasing.
so dp/dt>0.

But the rate of change of the gradient is decreasing so d2p/dt2 is less than zero.

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