What Is the Average Speed of a Rabbit in a 4 km Race?

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Homework Help Overview

The problem involves calculating the average speed of a rabbit participating in a 4 km race, where the rabbit runs a portion of the distance, takes a nap, and then runs the remainder at a faster speed. The participants are tasked with finding the average speed before the nap and the overall average speed.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the equations related to average speed and time, questioning the setup of the problem and the relationships between the speeds before and after the nap.

Discussion Status

Some participants have provided guidance on checking calculations and the correct addition of fractions. There is an ongoing exploration of the relationships between the speeds and the time taken for each segment of the race.

Contextual Notes

Participants are working under the constraints of homework rules and are attempting to reconcile their calculations with the expected answers from a textbook.

rash219
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Rabbits! Average Speed??

Homework Statement



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??

Homework Equations



Average Speed = Total Distance Traveled / Total Time Taken

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 can't 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 :confused:

Please Help...
 
Last edited:
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Check your working in solving the equation. You have made an error in adding the fractions.

Note that:
<br /> \frac{a}{b} + \frac{x}{y} \ne \frac{{a + x}}{{b + y}}<br />
 
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...
 
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.
 
ok then what is wrong with my final answer...why is the book saying its 9.00km/hr
 
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)
 
Darn what a silly mistake >_<! Thanks a million...came out right!
 

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