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f.4 maths

[匿名]

2009-09-16 5:34 am

tom and mary are due south and due east of a bus stop respectively, and the distance between them is 60m. they walk towards the bus stop at the same time with their own constant speeds. after 6 seconds, they are both 27m away from the bus stop. if mary walks slower than tom by 2m/s, find their speeds

回答 (1)

用戶10012

2009-09-16 5:53 am

✔ 最佳答案

Let speed of Tom = s.
Speed of Mary = s - 2
Distance travelled by Tom in 6 seconds = 6s
Distance travelled by Mary in 6 seconds = 6(s - 2)
so original distance of Tom from bus stop = 6s + 27
original distance of Mary from bus stop = 6(s - 2) + 27.
By Pythagoras theorem,
[6s + 27]^2 + [6(s - 2) + 27]^2 = 60^2
[2s + 9]^2 + [2(s - 2) + 9]^2 = 20^2
4s^2 + 81 + 36s + 4s^2 + 25 + 20s = 400
8s^2 + 56s - 294 = 0
4s^2 + 28s - 147 = 0
(2s - 7)(2s + 21) = 0
s = 7/2 or - 21/2 (rej.)
so speed of Tom is 3.5 m/s
speed of Mary = 1.5 m/s.

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