Constant velocity question

1. Jun 2, 2007

Aerosion

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

Kind of wordy, please bear with me:

A hare and a tortoise compete in a rase over a course 1.00 km long. The tortoise crawls straight at its maximum speed of 0.200m/s. The hare runs 8.00m/s for 0.800 km and stops to tease the tortoise. How close to the goal can the hare let the tortoise approach before resuming the race?

Assume both animals move steadily at their respective speeds.

2. Relevant equations

xf=1/2(Vi + Vf)*t

3. The attempt at a solution

So I used the above equation. I figure, tehy have 0.200km left to go after the hare stops. So I transfer .200km to m, and plug the numbers into the above equation to find out how long it would take the rabbit to get to the finish from there. I did the same for the turtle.

200=1/2(0+8.00)*t t=50s (rabbit)
200=1/2(0+.200)*t t=2000s (turtle)

So I saw it was a difference of 1950, so I decided to use the same equation to see how many meters the turtle could cover in 1950s, which would be how far the rabbit could afford to have the turtle get.

xf = 1/2(0+0.200)1500s xf=195m = 0.195km

Doesn't look right. Any suggestions?

2. Jun 3, 2007

ice109

the equation you need is the position function

$$X =X_o +V_ot + \frac{1}{2}at^2$$

though you're not given acceleration for the rabbit so i don't know.

and the equation for the tortoise is

$$X = V_ot$$

because you don't mention that the tortoise stops.

3. Jun 3, 2007

Pythagorean

Only the rabbit has .2 km left. Your indicating that the turtle has .2km left when the rabbit does.

You want to find out how many s the rabbit will need to get to the finish line, then how many km away from the finish line that many seconds is for the turtle (based on his speed).

Then you have to find a difference between two positions. Which ones do you think?

4. Jun 3, 2007

Aerosion

Oh oh oh...okay, I already have how many seconds the rabbit needs to get to the finish line in .2km (50s, from above).

I should use xf=(1/2)(Vi+Vf)t, right? (1/2)(0+0.200m/s)*50s = 5.00m.

This seems more correct. If it is, thanks.

5. Jun 3, 2007

ice109

the turtle never stops, his average velocity is the .2m/s not $$\frac{1}{2}(0+.200\frac{m}{s})$$