# physics

posted by .

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?

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 -

get your big ..

• physics -

thanks man.

### Answer This Question

 First Name School Subject Your Answer

### Related Questions

More Related Questions

Post a New Question