Find Distance

harpazo

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Molly drives to a destination at a rate of sixty miles per hour. She drives back over the same route at a rate of forty miles per hour due to traffic. If the round trip takes two hours, how far is the destination?

Solution:

D = rt

Going Rate = 60 mph
Returning Rate = 40 mph

Going:

D_1 = 1(60)

D_1 = 60 miles

Returning:

D_2 = 1(40)

D_2 = 40 miles

D_1 + D_2 = answer.

Yes? No?
 

MarkFL

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No...it appears you are assuming she takes 1 hour going there and 1 hour coming back. I would let \(t\) be the time it took to get there and so \(2-t\) will be the time it takes to get back...:

\(\displaystyle d=60t=40(2-t)\implies t=\frac{4}{5}\implies d=48\text{ mi}\)
 
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harpazo

Pure Mathematics
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No...it appears you are assuming she takes 1 hour going there and 1 hour coming back. I would let \(t\) be the time it took to get there and so \(2-t\) will be the time it takes to get back...:

\(\displaystyle d=60t=40(2-t)\implies t=\frac{4}{5}\implies d=48\text{ mi}\)
What's wrong with my reasoning?
 

MarkFL

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harpazo

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Math is very humbling. This is a middle school math question.
 

MarkFL

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We know distance is speed times time:

\(\displaystyle d=vt\)

Now suppose we are given two different speeds for the same distance:

\(\displaystyle d=v_1t_1=v_2t_2\)

If we assume \(\displaystyle t_1=t_2\) then it follows that we must also have \(\displaystyle v_1=v_2\). So, when we travel a certain distance, if speed changes, then so must the time it takes to cover that distance. If we double our speed, then the time it takes to cover the distance is cut in half, etc.
 
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harpazo

Pure Mathematics
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We know distance is speed times time:

\(\displaystyle d=vt\)

Now suppose we are given two different speeds for the same distance:

\(\displaystyle d=v_1t_1=v_2t_2\)

If we assume \(\displaystyle t_1=t_2\) then it follows that we must also have \(\displaystyle v_1=v_2\). So, when we travel a certain distance, if speed changes, then so must the time it takes to cover that distance. If we double our speed, then the time it takes to cover the distance is cut in half, etc.
Can you post and solve a word problem to show this fact?