# Homework Help: Intro Physics Help

1. Sep 9, 2007

### perfect_piccolo

A tortoise can run with a speed of 10.3 cm/s, and a hare can run 19.7 times as fast. In a race, they both start at the same time, but the hare stops to rest for 119 seconds. The tortoise wins by a shell (18.2 cm). How long does the race take?

2. Relevant equations

The only thing we've even covered so far is speed = d/t and v=delta x/delta t

3. The attempt at a solution

This is my first time ever taking a physics course and I just don't seem to be getting it. Can anyone help me set up some sort of equation that I would use to solve this problem? I've tried setting up two equations and then using substitution to try to solve for t, but my answers seem to come out to negative numbers. Am I even on the right track with this? Thanks so much

2. Sep 9, 2007

### learningphysics

Can you write out the equations you used and show how you tried to work the problem? We will guide you along in the process...

3. Sep 9, 2007

### perfect_piccolo

what I tried to do was set up two equations so that I could end up solving for t(time)

First I figured out that if the hare was going 19.7 times faster, he would be going 202.9 cm/s

so I had d total = 18.2 cm + d one and speed = d/t = 202.9 = d/t, so d=202.9t

then

speed = (18.2 cm + d one)/ t

10.3= (18.2 cm + 202.9t)/t

10.3 t = 18.2 cm + 202.9 t

t = -0.09

Which of course doesn't make sense. I don't think I'm using the right equations in the beginning of the problem.

4. Sep 9, 2007

### learningphysics

You're almost there... the only thing you missed was that if the turtle races for t seconds.... hare only races for t - 119.

So everything you did looks right except done = 202.9(t -119)

Only one suggestion I'd make is labelling your variables a little differently... like dhare (distance travelled by hare)... speedturtle... just to make sure that you don't get the turtle and hare variables mixed up.

5. Sep 12, 2007

### perfect_piccolo

Hey thanks so much, I finally got the right answer! :D

I'm having trouble with this one too...

Runner A is initially 3.8 mi west of a flagpole and is running with a constant velocity of 5.96 mi/h due east. Runner B is initially 2.94 mi east of the flagpole and is running with a constant velocity of 4.92 mi.h due west. How far are the runners from the flagpole when they meet? (West is positive and east is negative)

So what I've done is this:

d (total) = 6.74 - d

and

4.92 = d/t
4.92t=d
t=d/4.92

so then:

5.96=6.74 - d / t
5.96 = 6.74 - d / (d / 4.92)
5.96d + 1.2113=6.74-d
6.96d=5.527
d=0.794

(Which of course is wrong...although I'm getting pretty good at getting the wrong answer)

Can you see where I'm going wrong at...

Thanks!

6. Sep 12, 2007

### learningphysics

Cool! no prob.

this isn't the time that they meet. If the other runner was at rest, then it would be... but since both runners are moving towards each other, the time will be less than this.

I recommend writing the equation for the position of each of the two runners relative to the flagpole. Take west as negative... east as positive... so at t = 0, position of runner A is -3.8. at t=0, position of runner B is 2.94

write the equations for position of each of the two runners at any time t.

don't worry about getting the wrong answer... part of the learning process... all these problems take experience... making mistakes etc...

7. Sep 12, 2007

### perfect_piccolo

hmmmm I'm not sure if I understand what you mean here:

I recommend writing the equation for the position of each of the two runners relative to the flagpole. Take west as negative... east as positive... so at t = 0, position of runner A is -3.8. at t=0, position of runner B is 2.94

write the equations for position of each of the two runners at any time t.

====

Does that mean that for runner a t=d-3.8 mi/5.96 mi/h and for runner b t - d+2.94 mi/ 4.92 mi/h, or am I way off here?

8. Sep 12, 2007

### learningphysics

Oops... I apologize... your original work is right... I misunderstood... So d is the distance travelled by the 4.92 runner... and 6.74 - d is the distance travelled by the 5.96 runner.

I think your mistake was just algebra:

You should have parentheses here:
5.96=(6.74 - d)/ t
5.96 = (6.74 - d)/(d/4.92)

I don't understand how you get this next line:
5.96d + 1.2113=6.74-d

Last edited: Sep 12, 2007
9. Sep 13, 2007

### perfect_piccolo

Hey I don't know how I got it either lol

BUT

I did work out the right answer

To get it, you had to let d(1) = x, and d2 = 6.74-x (6.74 was the total distance between the two runners)

So now you could set up two equations:

s=d/t
5.96=x/t
x=5.96t

s=d/t
4.92=6.74-x/t

And then substitute:

4.92 = 6.74-5.96t/t
4.92t = 6.74-5.96t
10.88t = 6.74
t = 0.62

Now that you know the time, you can go:

5.96 = d/t
5.96t=d
5.96(.62) = d
3.70=d

So the last step you had to do was subtract 3.7 from 3.8 mi (where runner A started) and you come up with where they meet in relation to the flag pole, 0.10 mi.

Ta Da!

10. Sep 13, 2007

### learningphysics

looks good! just one thing... i may seem like a stickler... but use parentheses for

4.92=6.74-x/t ie:

4.92=(6.74-x)/t

on the message boards... because 4.92=6.74-x/t seems like 4.92=6.74-(x/t)