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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Stick on February 07, 2012, 05:44:56 pm

Title: Help!
Post by: Stick on February 07, 2012, 05:44:56 pm: I'm stuck on this question:

A biathlon event involves running and cycling. Kim can cycle 30km/h faster than she can run. If Kim spends 48 minutes running and a third as much time again cycling in an event that covers 60km, how fast can she run?

Thanks for your help! :)
Title: Re: Help!
Post by: b^3 on February 07, 2012, 06:31:31 pm: 1) $S_{C}=S_{R}+30$ .....[1]
$T_{R}=\frac{48}{60}=\frac{4}{5}$
$T_{C}=\frac{1}{3}T=\frac{4}{3}*\frac{4}{5}=\frac{16}{15}\: hrs$
$D_{C}+D_{R}=60$
Now remember that $Speed=\frac{distance}{time}$
So sub into [1] $S_{C}=\frac{D_{C}}{T_{C}}$
$S_{C}=\frac{60-D_{R}}{\frac{16}{15}}$
$S_{C}=\frac{15}{16}(60-D_{R})$
$\frac{15}{4}(60-D_{R})=S_{R}+30$ .....[1a]
Now $D_{R}=S_{R}*T_{R}=\frac{4}{5}S_{R}$
Sub into [1a] $\frac{15}{16}(60-\frac{4}{5}S_{R})=S_{R}+30$
$900-12S_{R}=16S_{R}+480$
$28S_{R}=420$
$S_{R}=15\: Km/H$

$\therefore\: Kim\: runs\: at\:15\: km/h$

Just remember its all about manipulating the speed equation and finding a couple of formulas $\mathbf{Speed=\frac{Distance}{Time}}$ :)

EDIT: FIXED IT
Title: Re: Help!
Post by: sam_i_am on February 07, 2012, 07:07:29 pm: to solve your problem you have to put it in an equation which is 60= 48/60 *x + 16/60 * (x+30) then you times it by 60 to get rid of the fraction,which makes it 3600=48*x+16*(x+30) then expand bracket, 3600 = 48*x+16*x+480 put like terms together, 3600= 64*x+480 takeaway 480, 3120= 64*x divide 64 which is 48.75 =x
Title: Re: Help!
Post by: Stick on February 07, 2012, 07:14:13 pm: Both are wrong. Sorry.
Title: Re: Help!
Post by: b^3 on February 07, 2012, 07:38:24 pm: Quote from: Stick on February 07, 2012, 05:44:56 pm
I'm stuck on this question:

A biathlon event involves running and cycling. Kim can cycle 30km/h faster than she can run. If Kim spends 48 minutes running and a third as much time again cycling in an event that covers 60km, how fast can she run?

Thanks for your help! :)
Ok I've fixed it in the post above, as the ratio between the time taken to cycle to the time taken to run should have been $\frac{4}{3}$ instead of $\frac{1}{3}$ as it is "a third as much time again" $(\frac{1}{3}+1=\frac{4}{3})$

Also note that the other method which sam_i_am outlined is quicker and more efficient when the right values are put in. :)
Title: Re: Help!
Post by: Stick on February 07, 2012, 08:30:36 pm: Thank you! :D