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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE General & Further Mathematics => Topic started by: 99.96 on November 07, 2011, 02:43:03 pm

Title: Graphs and Relations Last Q
Post by: 99.96 on November 07, 2011, 02:43:03 pm
For the last question, (how long can they talk to each other on the radio)
What did everyone get?
I got 1.38 hours, but i dont think ill get the marks for it anyway cos i didnt indicate that it was the answer.
Title: Re: Graphs and Relations Last Q
Post by: AnonymousLover on November 07, 2011, 03:15:27 pm
I got 2 hours?
Title: Re: Graphs and Relations Last Q
Post by: alpiano69 on November 07, 2011, 03:16:33 pm
1.375
Title: Re: Graphs and Relations Last Q
Post by: M2Char on November 07, 2011, 03:18:04 pm
1.375

This. Rounded off to 2dp, 1.38.
Title: Re: Graphs and Relations Last Q
Post by: Haras_Etak on November 07, 2011, 03:29:34 pm
i got 5.33 hours. I'm beginning to think this is wrong...
Title: Re: Graphs and Relations Last Q
Post by: Furbob on November 07, 2011, 03:30:20 pm
how did you guys work it out? I just went through substituting when t=1 up until t=7 and just counted which hours within the 3km distance (which was 2 hours) but knowing that the question asked for 2 dp im pretty sure im wrong
Title: Re: Graphs and Relations Last Q
Post by: Haras_Etak on November 07, 2011, 03:33:14 pm
I subtracted the girls equation from the other one and put it to equals less than 3. That meant t>5/3. Then i subtracted that from 7 but I think I just should have left it
Title: Re: Graphs and Relations Last Q
Post by: 99.96 on November 07, 2011, 03:36:09 pm
i did distance of girl - distance of guy = 3
found t for that which was 1.625
3-1.625 = 1.375
Title: Re: Graphs and Relations Last Q
Post by: tea.squaredd on November 07, 2011, 03:36:20 pm
I subtracted the girls equation from the other one and put it to equals less than 3. That meant t>5/3. Then i subtracted that from 7 but I think I just should have left it

That's what I did. However, i think we forget that one of the equations is only HALF the distance of the Guy. So some manipulation with the other half he travels is needed. Fml.
Title: Re: Graphs and Relations Last Q
Post by: M2Char on November 07, 2011, 03:37:35 pm
how did you guys work it out? I just went through substituting when t=1 up until t=7 and just counted which hours within the 3km distance (which was 2 hours) but knowing that the question asked for 2 dp im pretty sure im wrong

Find the first boundary by making the girl's equation equal the equation of the first part of the guy's hike plus 3.
Solve for t.
t = 1.625

Find the second boundary by making the girl's equation equal the equation of the second part of the guy's hike minus 3.
Solve for t.
t = 3

The time between those boundaries is the time that they are within 3km of each other.

3 - 1.625 = 1.375 ~= 1.38 hours
Title: Re: Graphs and Relations Last Q
Post by: davidle_10 on November 07, 2011, 04:43:51 pm
An easier way to think about it is to actually draw the graph of d=16-3t on the same axis.
Title: Re: Graphs and Relations Last Q
Post by: xdecay on November 07, 2011, 07:57:14 pm
i did distance of girl - distance of guy = 3
found t for that which was 1.625
3-1.625 = 1.375

sorry why did you have to minus 1.625 from 3?
Title: Re: Graphs and Relations Last Q
Post by: some dude on November 07, 2011, 08:34:05 pm
i did distance of girl - distance of guy = 3
found t for that which was 1.625
3-1.625 = 1.375

sorry why did you have to minus 1.625 from 3?

because the walkie talkie was in range at 1.625 hours and went out of range after 3 hours. so total hours is 3-1.625 =1.38
its easier to understand when you look at the graph
Title: Re: Graphs and Relations Last Q
Post by: some dude on November 09, 2011, 11:27:21 pm
I got 1.38 hours, but i dont think ill get the marks for it anyway cos i didnt indicate that it was the answer.

what do you mean indicate? as long as its the last number you write isnt that enough?