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September 19, 2014

September 19, 2014

Posted by **Anonymous** on Tuesday, October 24, 2006 at 2:18am.

Two students walk in the same direction along a straight path, at a constant speed- one at 0.90 m/s and the other at 1.90 m/s.

a) Assuming that they start at the same point and the same time, how much sooner does the faster student arrive at 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?

I calculated part A and got 456.2 seconds, but I am unsure how to do part B.

I am wondering what you did on A, if you can't do b?

Let me see your thinking on A.

I divided 780/.90 m/s which is 866.7 and then I divided 780/1.9 m/s which is 410.5 and then subtracted 866.7-410.5 to get 456.2 seconds.

This is a two-part question, and I got the first part, but I do not know how to do the second part.

Two students walk in the same direction along a straight path, at a constant speed- one at 0.90 m/s and the other at 1.90 m/s.

a) Assuming that they start at the same point and the same time, how much sooner does the faster student arrive at 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?

Distance1=speed1*time1

distance2=speed2*time2

subtract the second equation from the first.

distance1-distance2=speed1*time1 -speed2*time2

but the distances are the same.

0=1.9Time1-.90time2

But time2=time1*5.50min*60sec/min

Put that for time 2 in the equation, and solve for time1, and time2.

Now, you can solve for distance

distance=.90*time2

- physics -
**ohhjd**, Thursday, September 16, 2010 at 1:23amget your big ..

- physics -
**cool**, Wednesday, September 14, 2011 at 8:50pmthanks man.

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