SOLUTION: If I run 6 mph and my friend gives me a one hour head start. How long will I be running before my friend catches me?

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Question 1049446: If I run 6 mph and my friend gives me a one hour head start. How long will I be running before my friend catches me?
Found 2 solutions by advanced_Learner, MathTherapy:
Answer by advanced_Learner(501) About Me  (Show Source):
You can put this solution on YOUR website!
friend
time x%2B1
speed xmph
distance x%28x%2B1%29


you
time x
speed 6mph
distance 6x
x%28x%2B1%29=6x
%28x%5E2-5x%29=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-5x%2B0+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-5%29%5E2-4%2A1%2A0=25.

Discriminant d=25 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--5%2B-sqrt%28+25+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-5%29%2Bsqrt%28+25+%29%29%2F2%5C1+=+5
x%5B2%5D+=+%28-%28-5%29-sqrt%28+25+%29%29%2F2%5C1+=+0

Quadratic expression 1x%5E2%2B-5x%2B0 can be factored:
1x%5E2%2B-5x%2B0+=+1%28x-5%29%2A%28x-0%29
Again, the answer is: 5, 0. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-5%2Ax%2B0+%29

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

If I run 6 mph and my friend gives me a one hour head start. How long will I be running before my friend catches me?
Never will he/she catch you, if your friend's speed is: matrix%281%2C2%2C+%22+%22%3C=+6%2C+mph%29. 
However, if you stop running, he/she will catch you. Why do you ask such a question, anyway?
Have you read what you posted? Does it really make sense to you? I really hope it doesn't!