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September 19, 2014

September 19, 2014

Posted by **Margie** on Sunday, August 31, 2008 at 10:12pm.

a.) assuming that they start at the same point ant the same time, how much sooner does the faster arrive at a destination 780 m away?

b.) how far would the students have to walk so that the faster student arrives 5.50 min before the slower student?

- Physics -
**GK**, Monday, September 1, 2008 at 4:34pm(a) distance = speed x time

or time = distance/speed

The two times can be found by dividing the distance by each speed. Subtract the two times.

(b) Let t = time it takes the slower student.

5.50 min = 330 s

The distance is

0.90t = 1.90(t-330)

Solve for t, then substitute into 0.90t

- Physics -
**Paris**, Tuesday, September 16, 2014 at 5:18pmB. 570m

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