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Mathematics 15 Online
OpenStudy (anonymous):

John walks 1.20km north, then turns right and walks 1.20km east. His speed is 2.00m/s during the entire stroll If Jane starts at the same time and place as John, but walks in a straight line to the endpoint of John's stroll, at what speed should she walk if she wants to arrive at the endpoint just when John does?

OpenStudy (anonymous):

I get 1.414 meters/second.

OpenStudy (anonymous):

whats the formula you used?

OpenStudy (anonymous):

please explaine?

OpenStudy (anonymous):

He walked north 1200 meters at 2meters per second. Then he walked east 1200 meters at 2 meters per second. In the "parlance of physics" the magnitude of the square root of the sum of the squares is 1697 meters. That is the distance Jane must walk to catch up with John. John walked north for 600 seconds and walked east for 600 seconds for a total of 1200 seconds. velocity "v" = distance/time=1697m/1200s=1.414 m/s. \[\sqrt({1200^{2}}+1200^{2}) =1697\] \[1200m \div 2m/s=600 seconds\]

OpenStudy (anonymous):

cool thanks youve been a great help

OpenStudy (radar):

Good response, you've been fanned

OpenStudy (anonymous):

just glad to help!

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