anonymous
  • 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?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
I get 1.414 meters/second.
anonymous
  • anonymous
whats the formula you used?
anonymous
  • anonymous
please explaine?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • 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. \[\sqrt({1200^{2}}+1200^{2}) =1697\] \[1200m \div 2m/s=600 seconds\]
anonymous
  • anonymous
cool thanks youve been a great help
radar
  • radar
Good response, you've been fanned
anonymous
  • anonymous
just glad to help!

Looking for something else?

Not the answer you are looking for? Search for more explanations.