# Struggling with homework problem

1. Aug 24, 2009

### behnz06

1. The problem statement, all variables and given/known data

John walks 1.60 north, then turns right and walks 1.40 east. His speed is 1.80 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?

2. Relevant equations

3. The attempt at a solution
(((I figure out that the displacement is 2.13km. But I can't figure out how fast Jane needs to walk. I tried .9 but that's not the answer. My idea was that if you just divided John's time by half that would be the answer))))

2. Aug 24, 2009

### Staff: Mentor

Why by half?

What distance does she have to cover? How much time does she have to cover it? Calculate her speed.

3. Aug 24, 2009

### cerberus9

Well since you have John's displacement, it's going to be the same a Jane's, so you know that.

Then you also know that time has to be the same.

So solve for velocity

4. Aug 24, 2009

### behnz06

My thought process was that the triangle I made (using the 2 sides plus the displacement that I solved) could be seen as a square.lol. Thus, instead of going all around from ABC. AC would be a shortcut. I'm not quite sure if I was able to explain this clearly.

5. Aug 24, 2009

### cerberus9

it's not a square because John walks 1.60 km north and 1.40 km east. Those are different lengths of the sides, so it's not a square. but you still get a triangle when you put the two paths together haha.

6. Aug 24, 2009

### behnz06

By the way its 1.6 km and 1.4 km. And he moves at the rate of 1.8 m/s

7. Aug 24, 2009

### behnz06

I know it's not a square. But i believed it was a right angle. thus if I added a triangle just like that t would be a square

8. Aug 24, 2009

### cerberus9

well jane will be walking in a straight line, no?
so all you have to do is solve. you have the displacement and the time.

9. Aug 24, 2009

### behnz06

I don't know how to do that lol. would I set the problem up like 2.13km=1.8m/s over 3km?

10. Aug 24, 2009

### behnz06

Okay now i just tried converting 1.8miles pers second to km/s. Got 1800km/s and did 2.13 over 1800 km and got 1.18km/s

11. Aug 24, 2009

### behnz06

I mean 1.18 m per second

12. Aug 24, 2009

### cerberus9

well what equations are there that involve distance, velocity, and time?

13. Aug 24, 2009

### cerberus9

don't worry about converting things like that, it'll just confuse you tons more haha.

14. Aug 24, 2009

### behnz06

v=d/t

15. Aug 24, 2009

### behnz06

V= 2.13km/1.8mpersecond = 1.2 (wrong answer)

16. Aug 24, 2009

### cerberus9

1.8 m/s isn't a time, it's a rate. we're solving for a rate. that's John's rate, we don't want that. We want Jane's rate. So using the time and distance, what answer do you get?

17. Aug 24, 2009

### behnz06

I don't know. I got the problem wrong too many times. It says that the answer is 1.28m/s. I don't understand how they got it.

18. Aug 24, 2009

### ideasrule

What did you get for Jane's time? What did you get for Jane's displacement? Divide the second by the first.

EDIT: you got the right value for the displacement. You just need her time, which is the same as John's time. Find it and you're done.

19. Aug 24, 2009

### getitright

The answer is 1.27 m/s to reach distance (C) from starting point (A). You have most of the equation, just solve for distance over time.

Last edited: Aug 24, 2009