Race between a Speedy Tortoise and a Resting Hare

Click For Summary
SUMMARY

The race between a speedy tortoise and a resting hare illustrates the principles of relative motion and time management. The tortoise runs at 10 cm/s, while the hare runs at 200 cm/s but rests for 120 seconds. The tortoise wins the race by 20 cm, leading to the conclusion that the total time taken for the race is 240 seconds, and the length of the race is 2000 cm. This problem emphasizes the importance of setting up equations based on the conditions provided.

PREREQUISITES
  • Understanding of relative motion and speed calculations
  • Familiarity with basic algebra for solving equations
  • Knowledge of time-distance-speed relationships
  • Ability to set up and manipulate equations based on given conditions
NEXT STEPS
  • Study the concept of relative motion in physics
  • Learn how to derive equations from word problems
  • Practice solving problems involving time, distance, and speed
  • Explore more complex scenarios involving multiple moving objects
USEFUL FOR

Students studying physics, educators teaching motion concepts, and anyone interested in problem-solving techniques related to speed and distance calculations.

scottzilla
Messages
6
Reaction score
0

Homework Statement



Speedy tortoise can run 10cm/s, and a hare can run 20 times as fast. In a race, they both start at the same time, but the hare stops to rest for 2 minutes. The tortoise wins the race by a shell (20cm). A)How long does the race take? B)What is the length of the race.

Homework Equations





The Attempt at a Solution


For some reason I am lost on how to set up this problem. I keep wanting to use motion in 1 dimension with constant acceleration but I cannot come up with the answers given in the book. A step by step explanationj on how to solve this type of question would be awesome.

Scottzilla
 
Physics news on Phys.org
This question is probably a little simpler than you're trying to make it.

Okay, let's start by writing down what we know. Suppose that the length of the track is L and the time taken for the tortoise to complete that race, T, such that

v = \frac{dx}{dt}\Rightarrow 10 = \frac{L}{T}

Next we have the hare's information. We know that the hare stops of 2 minutes or 120 seconds, so the total time the hare is running is T-120. We also know that when the tortoise is at the finish line (x=L), the hare is 20cm behind him, i.e. x=L-20. Finally, we know that the hare can run at 20x10 cm/s. Hence,

v = \frac{dx}{dt}\Rightarrow 200 = \frac{L-20}{T-120}

Do you follow?
 
Ok I think I follow you. When I was trying to come up with my equations I was taking the hare's distance to be equal to the tortoise's distance minus 20cm. SO should I take those equations and solve one for (T)ime and then put that equation into the other one?
 
scottzilla said:
When I was trying to come up with my equations I was taking the hare's distance to be equal to the tortoise's distance minus 20cm.
That's exactly what I've done with my set of equations, L-20
scottzilla said:
SO should I take those equations and solve one for (T)ime and then put that equation into the other one?
Sounds good to me :approve:
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
1
Views
5K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
8K