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Gemma and Jack live at opposite ends of a street. Gemma had to deliver a parcel to Jack's home, Jack one to Gemma's home. They started their journeys at the same moment and each walked at constant speed, but not necessarily at the same speeds.

They first meet $c$ metres from Gemma's home and they meet again for a second time $(1000 - 2c)$ metres from Jack's home. You may assume they meet for the second time after both have delivered their parcels.

  1. Find how long their street is, leaving your answer in terms of $c$?
Harry B
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3 Answers3

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Let $x$ be Gemma's speed, $t$ be the time when she meets Jack the first time, $d$ is the total distance. Then $c=xt$. Note that together they covered the whole distance $d$. When they meet the second time, they cover distance $2d$ so it will take twice as long. Thus, $2t=\frac{d-c}{x}+\frac{1000-2c}{x}$. But $t=\frac{c}{x}$, therefore $$d-c+1000-2c=2c \implies d=5c-1000$$

Vasili
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I suggest you try the following: $$velocity=\frac{distance}{time}$$ or $$time=\frac{distance}{speed}$$ You need to write the distance each one travels in terms of $c$ and the length of the street $L$. For example, at the first intersection, Gemma travels $c$ meters, and Jack travels $L-c$. Since the time is the same: $$\frac{c}{v_G}=\frac{L-c}{v_J}$$You can write this in terms of ratio of the velocities:$$\frac{v_J}{v_G}=\frac{L-c}c$$ Do the same for the other intersection. Find the same ratio, and then you would get an equation containing only $L$ and $c$.

Andrei
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Let Gemma's place be at the origin of the $x$-axis, and let Gamma's speed be $v_1$ , and Jack's speed be $v_2$. And let $L$ be the length of the street between them. Then

$p(t) = v_1 t , q(t) = L - v_2 t $

At $t_1$ the first meeting time, we have

$t_1 = \dfrac{L}{v_1 + v_2} $

$p(t_1) = c = \dfrac{v_1 L}{ v_1 + v_2 } $

Hence,

$c = \dfrac{L}{ 1 + \dfrac{v_2}{v_1} }$

Now the distance travelled by Gemma plus the distance travelled by Jack between the first and second meetings is $ 2 L = (v_1 + v_2)(t_2 - t_1) $

We have to calculate $t_2$. The distance remaining for Gemma is $ L - c $ in the original direction, which will be traversed in $\Delta t_1 = \dfrac{L - c}{v_1}$. In addition, the time taken for the opposite direction of movement for Gemma is,

$\Delta t_2 = \dfrac{ 1000 - 2 c }{v_1} $

Hence $t_2 - t_1 = \dfrac{ L + 1000 - 3 c}{v_1}$

In this time period, the sum of distances of Gemma and Jack is $2 L$. Hence,

$2 L = \dfrac{(v_1 + v_2)}{v_1} ( L + 1000 - 3 c) $

$ 2 L = \dfrac{L}{c} ( L + 1000 - 3 c) $

$ 2 c = L + 1000 - 3 c $

$ L = 5 c - 1000 $

Hosam Hajeer
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