Calculating Average Speed for a Trip: A Physics Challenge

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To average 90 km/hr for a trip, if the first half is traveled at 48 km/hr, a significantly higher speed is required for the second half. The calculation shows that to meet the average speed goal, the second half must be completed in just 3.75 minutes, necessitating a speed of 720 km/hr. This is impractical and highlights the challenge of achieving the desired average speed after a slow first half. The discussion emphasizes the importance of understanding the relationship between distance, speed, and time in physics. Overall, the scenario illustrates the complexities of average speed calculations in real-world applications.
Niki Lively
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1. You plan a trip on which you want to average 90 km/hr. You cover the first half of the distance at an average speed of only 48 km/hr. What must your average speed be in the second half of the trip to meet your goal? Note that the velocities are based on half the distance, not half the time. Is this reasonable?


2. Not sure.



3. This is the beginning of something new. Our Physics class has started a couple weeks ago, so my fellow classmates & I are still settling into the new concept of our first Physics class; our first time allowed to think "outside the box". I'm very intrigued, but confused. As I said, this is something new in the chapter, & he hasn't given us many details on where to start. I just am confused with which sort of formula to use. I thought back to previous math, & I'm thinking along finding the median. :confused: So, I thought about what would work out to divide by 2 to get 90; obviously 180. 48-180=132. So, 132 km/hr would have to be the average speed for the second time...? I know I can't possibly be correct. It's probably something quite obvious that I'm just not catching. So, can someone help me by giving me the proper formula so I can figure it out? I'm just not sure on where to place what... I know conversion formulas, but that is irrelevant right now. Any help would be greatly appreciated. :smile:
 
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Niki Lively said:
1. You plan a trip on which you want to average 90 km/hr. You cover the first half of the distance at an average speed of only 48 km/hr. What must your average speed be in the second half of the trip to meet your goal? Note that the velocities are based on half the distance, not half the time. Is this reasonable?


2. Not sure.



3. This is the beginning of something new. Our Physics class has started a couple weeks ago, so my fellow classmates & I are still settling into the new concept of our first Physics class; our first time allowed to think "outside the box". I'm very intrigued, but confused. As I said, this is something new in the chapter, & he hasn't given us many details on where to start. I just am confused with which sort of formula to use. I thought back to previous math, & I'm thinking along finding the median. :confused: So, I thought about what would work out to divide by 2 to get 90; obviously 180. 48-180=132. So, 132 km/hr would have to be the average speed for the second time...? I know I can't possibly be correct. It's probably something quite obvious that I'm just not catching. So, can someone help me by giving me the proper formula so I can figure it out? I'm just not sure on where to place what... I know conversion formulas, but that is irrelevant right now. Any help would be greatly appreciated. :smile:



Sometimes you can get a feel for how a problem works if you make up a simple example. The equation that applies here is s = vt where s is the distance, v is the velocity and t is the time. Suppose, for example you wanted to go 90km. at an average speed of 90 km/hr that would take you 1 hour. The problem said you travel the first half of the trip at an average velocity of 48 km/hr. That means it took

t = \frac{s}{v} = \frac{45}{48} = \frac{15}{16} hour

To achieve your desired goal of averaging 90 km/hr for the whole trip you would have to travel the last 45 km in 1/16 of an hour or 3 and 3/4 minutes. I think you might get a speeding ticket!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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