Really simple math question embaressed I'm asking.

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RIPP_n_WIPE
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Really simple math question embaressed I'm asking.

Post by RIPP_n_WIPE »

Okay. I'm finally getting my feet wet again mathematically. I've been out of the loop for about a year and a half and am surprised about how much I've lost.

So here's the question.

Alice and Bob are going from their home to a friend’s house, traveling by foot. Alice leaves home first and strolls slowly at a rate of 2 mph towards the friend’s house. Twelve minutes later (1/5 hour later), Bob leaves home and jogs towards the friend’s house at 5 mph.

(a) When will Bob catch up to Alice?
(State your final answer carefully as either “Bob will catch up to Alice ___ minutes after Bob leaves home” or “Bob will catch up to Alice ___ minutes after Alice leaves home”)

(b) How far has each person traveled at this point?

I dunno exactly what I'm doing wrong with it. I refuse to guess and check for the answer.

Thus far I've set up the two like my book had a similar question set up. Where Alices distance = d and her rate is 2 and her time being t+1/5 getting me 2t+2/5. Bobs distance = d his rate = 5 and his time t yielding me 5t. When I set up the equality of 5t = 2t+2/5 gets me t=2/15 which I know is off because Alice is already gone for 1/5 of an hour,

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General Soontir Fel
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Re: Really simple math question embaressed I'm asking.

Post by General Soontir Fel »

You solved everything correctly. The equations solve to t=2/15, where t is the time since Bob left the house. So Bob will catch up to Alice 2/15 of an hour, or 8 minutes after leaving the house, which is 20 minutes after Alice left. Since Bob jogs at 5 mph, it will be (2*5)/15, or 2/3 of a mile from the house.
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RIPP_n_WIPE
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Re: Really simple math question embaressed I'm asking.

Post by RIPP_n_WIPE »

Wow thanks.

I am the hammer, I am the right hand of my Lord. The instrument of His will and the gauntlet about His fist. The tip of His spear, the edge of His sword. I am His wrath just as he is my shield. I am the bane of His foes and the woe of the treacherous. I am the end.


-Ravus Ordo Militis

"Fear and ignorance claim the unwary and the incomplete. The wise man may flinch away from their embrace if he girds his soul with the armour of contempt."
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Re: Really simple math question embaressed I'm asking.

Post by Surlethe »

B = 5t
A = 2(t + 0.2)

Time of intersection: A = B so 5t = 2(t + 0.2) implies 3t = 0.4, so t = 2/15 hr = 8 min. Distance of intersection: 5*2/15 mi = 10/15 mi = 2/3 mi.

The conceptual part is probably the most difficult: set your origin when \textit{Bob} leaves the house. Then Alice's equation of motion is A = 2(t+0.2): since she's already traveled for 12 minutes = 0.2 hr, she's gone 2*0.2 miles. You can set the origin when Alice leaves the house, but then you have to deal with a bent function when Bob starts to jog.
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Re: Really simple math question embaressed I'm asking.

Post by Singular Intellect »

Was it silly for me to immediately be suspicious of the question, since there was no stated distance between the houses in question, thus Alice could've arrived at it well before Bob left and any interception point was based purely on a unsubstantiated assumption?
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Re: Really simple math question embaressed I'm asking.

Post by CaptainChewbacca »

Singular Intellect wrote:Was it silly for me to immediately be suspicious of the question, since there was no stated distance between the houses in question, thus Alice could've arrived at it well before Bob left and any interception point was based purely on a unsubstantiated assumption?
For the question to be answerable, you have to assume they will meet before they arrive at the friend's house. I'm sure the friend could be next door, but there's no way to do that.
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