# Homework Help: Rabbits Average Speed?

1. Jun 23, 2007

### rash219

Rabbits!!!! Average Speed??

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

A rabbit takes part in a 4.00km race. First the rabbit runs 0.500 km then stops and takes a 90.0min nap. when he wakes up he runs twice as fast completing the race in a total time of 1.75h.

A:-Calculate the average speed of the rabbit?
B:-Calculate the average speed of the rabbit before he stopped for a nap??

2. Relevant equations

Average Speed = Total Distance Traveled / Total Time Taken

3. The attempt at a solution

I got the 1st one, A, the answer to that was 2.29km/h......

Ok i have been trying to sole the 2nd part of this myth but i still cant figure out how to get the 9.00 km/hr.. my attempt is below

Converting 90 min to hrs = 1.5h...therefore actual time taken for the 2 distances would = to 0.25h ie 1.75h - 1.5h

Thus forming the 1st equation T1 + T2 = 0.25h --------(1)

v = avg. speed; d = distance; t = time

now we know that v1 = d1 / t1 and v2 = d2 / t2......
where d1 = 0.5km and d2 = 3.5km
and according to the question we can form another equation stating that

2*v1 = v2 -----------------(2)

using the given formula we can write an equation like

(d1 / v1) + (d2 / v2) = t1 + t2

(0.5km / v1) + (3.5km / 2*v1) = 0.25h

(4.0km / 3*v1) = 0.25h

3*0.25h*v1 = 4.0km

therefore v1 = 4.0km / 0.75h = 5.34 km/hr...........

WHAT AM DOING WRONG?????????

according to my book the answer is 9.00km/h

Last edited: Jun 23, 2007
2. Jun 24, 2007

### danago

Note that:
$$\frac{a}{b} + \frac{x}{y} \ne \frac{{a + x}}{{b + y}}$$

3. Jun 24, 2007

### rash219

But according to the question the rabbit runs twice as fast so isn't it correct to say that 2 times V1 would equal to V2 !!! therefore substituting it in the problem..where v2 = 2 * v1...

4. Jun 24, 2007

### danago

yes that part is right, so this is correct:

(0.5km / v1) + (3.5km / 2*v1) = 0.25h

But remeber, when you add fractions, you need to find a common denominator.

5. Jun 24, 2007

### rash219

ok then what is wrong with my final answer.....why is the book saying its 9.00km/hr

6. Jun 24, 2007

### danago

You didnt add the fractions correctly.

(0.5km / v1) + (3.5km / 2*v1) $$\ne$$ (4.0km / 3*v1)

But rather:

(0.5km / v1) + (3.5km / 2*v1) $$=$$ (4.5km / 2*v1)

7. Jun 24, 2007

### rash219

Darn what a silly mistake >_<! Thanks a million..........came out right!!!