Motion in One Dimension: The Tortoise and The Hare Race

In summary, a turtle and a rabbit have a footrace over a distance of 4.00km. The rabbit runs 0.500km and then takes a 90-minute break before continuing at twice his original speed. He finishes the race in a total time of 1.75 hours, winning the race. To find the average speed of the rabbit, the distance and time for the entire race are used. The average speed is 2.29km/h. To find the rabbit's average speed before the break, the distance and time for the first part of the race are used, but the time is adjusted to account for the break. The resulting average speed is 9km/h.
  • #1
DracoMalfoy
88
4

Homework Statement



A turtle and a rabbit engage in a footrace over a distance of 4.00km. The rabbit runs 0.500km and then stops for a 90min nap. Upon awakening, he remembers the race and runs twice as fast. Finishing the course in a total time of 1.75h, the rabbit wins the race.

A) Calculate the average speed of the Rabbit.

B) What was his average speed before he stopped for a nap? Assume no detours or doubling back.

The question is asking to find the answers in km/h

Homework Equations


  • Conversion
  • Average Velocity: Δd/Δt= df-di/tf-ti
  • T=D/T ; D=V⋅T

The Attempt at a Solution



1. I first labeled values.
[/B]
  • Trip1
Δd: 0.500km
AvgV: ?
T: ? = 0.25h
  • Break
T: 90.0min= 1.5h

  • Trip2
2v1: ?= 4.58km/h

  • Trip Total
Δd: 4.00km
T: 1.75h
AvgV: 2.29km/h

2. I then solved for the average speed of the Rabbit:

Average Velocity: Δd/Δt= 4.00km/1.75h = 2.29km/h
I'm guessing that if the rabbit was running twice the avg speed in trip 2, I would have to multiply 2.29km/h by 2 for the velocity of trip two.

3. Next, I converted the Rabbits break time: 90.0mins to hours and got 1.5 hours

4. Since the next step is asking for the average speed BEFORE the rabbit stopped to take a nap, it requires more steps.
~
So I subtracted the rabbits total time by the break time and got: ΔT=0.25h, or 15 minutes. The rabbit ran 15 minutes before taking a break.
~I then subtracted the distances of the total and trip 1 and Δd= 3.50km

*This is where I got stuck. I'm not sure what to do next. I know that I'm supposed to calculate the average speed, so would that mean that I'm supposed to subtract the distances, divide by 0.25h, then subtract that average speed from the doubled speed of trip 2?

In that case, my answer was 9.5km/hr. But the book states that the answer is 9.0km/hr. What did I do wrong here?
 
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  • #2
DracoMalfoy said:

Homework Statement



A turtle and a rabbit engage in a footrace over a distance of 4.00km. The rabbit runs 0.500km and then stops for a 90min nap. Upon awakening, he remembers the race and runs twice as fast. Finishing the course in a total time of 1.75h, the rabbit wins the race.

A) Calculate the average speed of the Rabbit.

B) What was his average speed before he stopped for a nap? Assume no detours or doubling back.

The question is asking to find the answers in km/h

Homework Equations


  • Conversion
  • Average Velocity: Δd/Δt= df-di/tf-ti
  • T=D/T ; D=V⋅T

The Attempt at a Solution



1. I first labeled values.
[/B]
  • Trip1
Δd: 0.500km
AvgV: ?
T: ? = 0.25h
  • Break
T: 90.0min= 1.5h

  • Trip2
2v1: ?= 4.58km/h

  • Trip Total
Δd: 4.00km
T: 1.75h
AvgV: 2.29km/h

2. I then solved for the average speed of the Rabbit:

Average Velocity: Δd/Δt= 4.00km/1.75h = 2.29km/h
I'm guessing that if the rabbit was running twice the avg speed in trip 2, I would have to multiply 2.29km/h by 2 for the velocity of trip two.

3. Next, I converted the Rabbits break time: 90.0mins to hours and got 1.5 hours

4. Since the next step is asking for the average speed BEFORE the rabbit stopped to take a nap, it requires more steps.
~
So I subtracted the rabbits total time by the break time and got: ΔT=0.25h, or 15 minutes. The rabbit ran 15 minutes before taking a break.
~I then subtracted the distances of the total and trip 1 and Δd= 3.50km

*This is where I got stuck. I'm not sure what to do next. I know that I'm supposed to calculate the average speed, so would that mean that I'm supposed to subtract the distances, divide by 0.25h, then subtract that average speed from the doubled speed of trip 2?

In that case, my answer was 9.5km/hr. But the book states that the answer is 9.0km/hr. What did I do wrong here?

Please turn off the bold font. It is very distracting, and comes across like you are yelling at us. Bold section headings are OK, but bold equations, etc., are really, really annoying.

Anyway, the rabbit does not run for 15 minutes before taking a nap; his total running time (before and after the nap) is ##t_R = 1.75 - 1.5 = 0.25 ##hr. That is split into two parts: ##t_{R1}## before his nap and ##t_{R2}= t_R - t_{R1}## after his nap. He runs at speed ##v## over the first part and at speed ##2v## over the second part. Knowing both the distance run and the total running time, you can get enough equations to determine both ##t_{R1}## and ##v##.
 
  • #3
Ray Vickson said:
Please turn off the bold font. It is very distracting, and comes across like you are yelling at us. Bold section headings are OK, but bold equations, etc., are really, really annoying.

Anyway, the rabbit does not run for 15 minutes before taking a nap; his total running time (before and after the nap) is ##t_R = 1.75 - 1.5 = 0.25 ##hr. That is split into two parts: ##t_{R1}## before his nap and ##t_{R2}= t_R - t_{R1}## after his nap. He runs at speed ##v## over the first part and at speed ##2v## over the second part. Knowing both the distance run and the total running time, you can get enough equations to determine both ##t_{R1}## and ##v##.

Sorry about the bold font. It makes me organize the information better sometimes. I am a bit confused now... I am not sure where to start now after finding the first part. Can you explain it a different way?
 
  • #4
DracoMalfoy said:
Sorry about the bold font. It makes me organize the information better sometimes. I am a bit confused now... I am not sure where to start now after finding the first part. Can you explain it a different way?

For the first part of the question you don’t know the times. You can’t solve it by assigning times to the pieces. What you know is the distances of the pieces, the total time, and a relation between the velocities of the two pieces. Subtract the nap time and you will have the total time for the two pieces. You will need to write an expression with the initial speed as an unknown relating the two distances to the total time and solve for the speed.
 
  • #5
DracoMalfoy said:
Sorry about the bold font. It makes me organize the information better sometimes. I am a bit confused now... I am not sure where to start now after finding the first part. Can you explain it a different way?

You do not know the speed ##v## over the first part of the run, but if you did know it you could write an expression for the running-time of the first part (before the nap). What is that expression? You do not know the speed over the second part of the run (after the nap), but you know it is ##2v##. In terms of the unknown ##v##, what is the expression for the running-time of the second part? You know the total running time (first+second part), so you can write an equation involving ##v##.
 
  • #6
Ray Vickson said:
You do not know the speed ##v## over the first part of the run, but if you did know it you could write an expression for the running-time of the first part (before the nap). What is that expression? You do not know the speed over the second part of the run (after the nap), but you know it is ##2v##. In terms of the unknown ##v##, what is the expression for the running-time of the second part? You know the total running time (first+second part), so you can write an equation involving ##v##.

t_total=(d_beforenap /v_beforenap)+(d_afternap/(2*v_beforenap))?
 
  • #7
DracoMalfoy said:
t_total=(d_beforenap /v_beforenap)+(d_afternap/(2*v_beforenap))?
You tell me.
 
  • #8
I'm honestly about to give up. This is really frustrating me. I can't even think clearly anymore.
 
  • #9
DracoMalfoy said:
I'm honestly about to give up. This is really frustrating me. I can't even think clearly anymore.

Go for a walk; take a break. Go out with some friends.

Come back when you feel a bit refreshed. Maybe leave it overnight and return to it after a good night's sleep.

You really do have everything you need; it is just a matter of having some confidence in your own abilities, and carrying through from start to finish---right or wrong.
 
  • #10
Let v be the speed of the rabbit during the first 0.5 km, and let 2v be the speed of the rabbit during the remaining 3.5 km. Algebraically, in terms of v, how much time ##t_1## does it take for the rabbit to run the first 0.5 km? Algebraically in terms of v, how much time ##t_2## does it take for the rabbit to run the final 3.5 km? Including the 90 minute nap, algebraically in terms of v, how much time does it take for the rabbit to run the entire race? This should also be equal to 1.75 hr. What value of v is required for the algebraic expression for the total time to match the 1.75 hr?
 
  • #11
0.25h= 0.5km/s1+ 3.5km/2s1

0.25h= (2)0.5km/(2)s1+ 3.5km/2s1

0.25h= 1km/2s1 + 3.5/2s1

0.25h= 4.5km/2s1

0.25h(2s1)= 4.5km

2s1/(0.25h)= 4.5km/(0.25h)

2s1= 18km/h

s1= 9km/h
 

FAQ: Motion in One Dimension: The Tortoise and The Hare Race

1. What is motion in one dimension?

Motion in one dimension is the movement of an object along a single axis, such as a straight line, without any change in direction.

2. What is the difference between speed and velocity?

Speed refers to how fast an object is moving, while velocity takes into account the direction of the object's motion.

3. How does the story of the tortoise and the hare demonstrate the concept of motion in one dimension?

In the story, the tortoise and the hare race in a straight line from start to finish, representing motion in one dimension. The hare's faster speed and varying velocity demonstrate the difference between the two concepts.

4. Can an object have a constant speed but changing velocity?

Yes, an object can have a constant speed but changing velocity if it is changing direction. For example, a car going around a circular track at a constant speed will have a changing velocity because it is constantly changing direction.

5. How can we calculate an object's average speed?

The average speed of an object is calculated by dividing the total distance traveled by the total time taken.

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