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Question #fe8f6

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Answer 1 · 2018-02-12T23:52:43

Jeffrey and Marcus would have passed each other after $2.5$ $2.5$ hours at 11:45 AM.

Explanation:

One way to picture this is by looking at the distance between Jeffrey and Marcus at some elapsed time in hours, $t$ $t$ .

Start with 410 (km).

$410$ $410$

For every hour that passes, Jeffrey is closing this distance between him an Marcus by $80$ $color(red)80$ km/h.

$410 - 80 t$ $410 - color(red)80t$

On the other side, however, Marcus is closing this distance between himself and Jeffrey by $120$ $color(blue)120$ km/h, but only after 45 minutes, or $\frac{3}{4}$ $color(green)(3/4)$ hours.

$410 - 80 t - 120 (t - \frac{3}{4})$ $410 - color(red)80t - color(blue)120(t - color(green)(3/4))$

This may also be thought of as "offsetting" the time Marcus has spent travelling by 45 minutes.

Now, the question has boiled down to finding a value of $t$ $t$ , for which the distance between Jeffrey and Marcus is $0$ $color(purple)0$ .

$0 = 410 - 80 t - 120 (t - \frac{3}{4})$ $color(purple)0 = 410 - color(red)80t - color(blue)120(t - color(green)(3/4))$

$0 = 410 - 80 t - 120 t + 90$ $0 = 410 - 80t - 120t + 90$

$0 = 500 - 200 t$ $0 = 500 - 200t$

$200 t = 500$ $200t = 500$

$t = \frac{5}{2}$ $t = 5/2$

Finally, we need to add this time to 9:15 AM.

$\frac{5}{2} = 2 \frac{1}{2}$ $5/2 = 2 1/2$

The time elapsed was $2 \frac{1}{2}$ $2 1/2$ hours or $2$ $2$ hours and $30$ $30$ minutes, so they passed each other at 11:45 AM.

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