- #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
AvgV: ?
T: ? = 0.25h
- Break
- Trip2
- Trip Total
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?