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homework help

W

washablemarker

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I need help with some homework...
Runner A is initially 6 km west of a flagpole and is running with a constant velocity of 9 km/hr due east. Runner B is initially 5 km west of the flagpole and is running with a constant velocity of 8 km/hr due west. What will be the distance of the two runners from te flagpole.
this might help: avg velocity = delta X / time , delta X = Xfinal - Xinitial , delta X = displacement (how far somethings moved) , the time it takes for the runners to meet is the same
thanks in advance, any help will really be appreciated
 
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washablemarker said:
Runner A is initially 6 km west of a flagpole and is running with a constant velocity of 9 km/hr due east. Runner B is initially 5 km west of the flagpole and is running with a constant velocity of 8 km/hr due west.

Shouldn't Runner B run from east to west?
 
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*Reads question* Argh this is a math question! *Starts running from point 'A' to point 'Ihatemaths'*
 
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For shorthand, I'll use v=velocity, dx=delta x, t=time, x0=initial position, xf=final position

OK, so you know v = dx/t, so it follows that dx=v*t and dx+x0=xf. What you need is an equation that incorporate all this. The following is true:

xf = v*t + x0

The value xf will contain the position on the axis of the runner after the specified time. Once you have found t, both values of xf will be the same, i.e. they will have met. Assuming the flagpole is on an axis, with east positive and west negative, the values from the original equation fit like so:

xf = 9 * t - 6
xf = -8 * t + 5

In other words, the first runner starts at point -6 (6 km west) and runs 9km/h east. The second runner starts at point 5 (5km east) and runs -8km/h, or 8km/h west. Solve this for t, then put t into each to make sure they match, and you'll have xf. The absolute value of xf will tell you distance, the sign will tell you direction.

edit: I assumed that the question was 'how far from the flagpole are they when they meet?' as this is quite a common question. If that's not the question, this should still help :)
 

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