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Two bike riders $X$ and $Y$ both start at 2 PM riding towards each other from $40$ km apart. $X$ rides at $30 \frac{\mathrm{km}}{\mathrm{h}}$, and $Y$ at $20 \frac{\mathrm{km}}{\mathrm{h}}$. If they meet after $t$ hours, find when and where they meet.

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At what speed are they approaching each other? Or in other words, at what rate is the distance between them shrinking? –  André Nicolas Jul 20 '13 at 0:34
    
That is unknown –  Mark Zakhem Jul 20 '13 at 0:35
    
It is known, $20+30$. –  André Nicolas Jul 20 '13 at 0:36
    
Wait actually, X is riding at 30km/h and Y at 20 km/h –  Mark Zakhem Jul 20 '13 at 0:36
    
would it be t(20+30)=40 –  Mark Zakhem Jul 20 '13 at 0:37

2 Answers 2

up vote 1 down vote accepted

Let $40 \text{km}= x + y$, where $x$ is trace of $X$ and $y$ trace of $Y$. So $y=40-x$.

From their speed we can say: $30t=x$ and $20t=y$, so $t=\frac x{30}$ and $t=\frac{40-x}{20}$.

Now just modify a little bit $\frac x{30}=\frac{40-x}{20}$.

$x/3=(40-x)/2$,

$2x=120-3x$

$5x=120$

$x=24$km

Ergo $y=16 $km and time is $t=0$,$8 $h, so they met after $48$ minutes, ($2:48$ pm).

(sorry for the style, just figuring everything out)

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Good Detailed Answer. Thanks! –  Mark Zakhem Jul 26 '13 at 8:24

The combined rate at which the two are moving toward each other is given by $20 + 30 = 50$km/hour. Put $t = \text{time in hours}$, and solve for $t$, given: $$t \text{ hrs}\times \frac{50\text{km}}{\text{hour}} = 40 \text{km}$$

Hence we have that $t = \dfrac {40}{50} = \dfrac 45 \;\text{hour}.\;$ Compute $t \times 60\frac{\text{min}}{\text{hr}} = \dfrac 45\times 60 = 48 \;\text{minutes}$ to obtain the time in minutes, which you can add to $2$:$00$ pm, giving us that X and Y meet at $2$:$48$ p.m.

That will give you when they meet. You can then use the original value of $t$ to compute where they meet. Can you figure out how to use the rate at which $X$ is riding to obtain the position at which $X$ meets $Y$? Let's call this point $M$.

Let's just do it: Put $X$'s starting position at $0$, with $X$ heading east, in the positive x-direction, and $Y$ at $40$km along the number line, heading "west" toward the origin. Then use your original $t$ (as a fraction of an hour) along with the rate at which $X$ is riding, $30$km/hour, to compute the distance from $0$ that $X$ travels in $t$ hours: $\text{distance} = t\times 30\text{km/hour} = \dfrac 45 \times 30 = 4\times 6 = 24 \text{km}$. This gives us the distance that $X$ travels until $X$ and $Y$ meet. So the value of $M$ will then be $X = 0 + d = 24$, and will give the point along the x-axis (or number-line) at which $X$ and $Y$ meet. Note, that means that $Y$ traveled $16$km.

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Nice detailed write up! +1 CONGRATS on over 2k answers! –  Amzoti Jul 20 '13 at 6:01
    
+70k ${}{}{}{}{}$ :-) –  B. S. Jul 20 '13 at 15:11
    
@Babak Not quite (but almost there!) –  amWhy Jul 20 '13 at 15:12
    
@amWhy: IT WILL BE :-) –  B. S. Jul 20 '13 at 15:13

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