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

Click For Summary
SUMMARY

The average speed of a rabbit in a 4 km race, after running 0.5 km and taking a 90-minute nap, is calculated to be 2.29 km/h overall. The rabbit runs twice as fast after the nap, leading to a calculated average speed before the nap of 9.00 km/h. The key equations used include Average Speed = Total Distance / Total Time and the relationship between the speeds before and after the nap, where v2 = 2 * v1. The error in the initial calculation stemmed from incorrect addition of fractions.

PREREQUISITES
  • Understanding of basic algebraic equations
  • Knowledge of average speed calculations
  • Familiarity with unit conversions (e.g., minutes to hours)
  • Ability to manipulate fractions and find common denominators
NEXT STEPS
  • Study the concept of average speed in physics
  • Learn how to solve equations involving fractions
  • Practice unit conversion techniques for time and distance
  • Explore more complex motion problems involving speed and time
USEFUL FOR

Students studying physics or mathematics, educators teaching speed and motion concepts, and anyone interested in solving real-world problems involving average speed calculations.

rash219
Messages
22
Reaction score
0
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:
Physics news on Phys.org
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!
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
8K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
28K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
8
Views
15K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
8K