
By: tchtby (offline) Friday, August 09 2013 @ 05:46 AM CDT (Read 1134 times)



tchtby 
pls help me solve these 2 qn
thks

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Registered: 10/15/12 Posts: 13





By: Lisa Wong (offline) Saturday, August 10 2013 @ 12:59 AM CDT



Lisa Wong 
Hope I'm of some help. Pls see attachment for solution. I'm also a P6 anxious mum.

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Registered: 05/09/10 Posts: 1





By: echeewh (offline) Saturday, August 10 2013 @ 02:43 PM CDT



echeewh 
Hey <tchtby>
Following is my worked solution for Q10:
For T, J to pass / meet each other (from the opp direction), the duration / time taken has to be the same.
Applying the <Common Time> method, where the Distance ratio is the same as the Speed ratio, we have ...
sT : sJ
160 : 200
4 : 5
dT : dJ
4 : 5
This distance of 9u (dT + dJ) is based on their initial starting points on the track (i.e. T and J were 200 m apart).
1u distance = 200/9 m
Based on the given circular track of 400m, T and J would meet for a 2nd time after running / completing a combined / joint distance (dT + dJ) of 600m.
dT = 4u × 3 = 12u
12u distance = 12 × (200/9) = 266 2/3 m
=========
Trust this helps.
Do let me know again if there's further clarification.
Cheers,
Edward

Active Member
Registered: 04/21/11 Posts: 627



