Calculating Passing Speed: Understanding the Concept and Formula Explained"

  • Thread starter Thread starter sandy
  • Start date Start date
Click For Summary
When two cars pass each other in opposite directions, their passing speed is the sum of their individual speeds. In this case, one car traveling at 100 km/h and another at 150 km/h results in a combined passing speed of 250 km/h. The concept of passing speed is equivalent to the rate of separation, as it reflects the relative speed of one car with respect to the other. Understanding this can be illustrated by considering the scenario where both cars are moving towards each other, leading to the addition of their speeds. Therefore, the passing speed is not zero but rather the total of both speeds combined.
sandy
Messages
1
Reaction score
0
Can someone please help:

Question: 2 cars are passing each other on opposite directions. One at 100 Km and the other at 150 Km. What is the passing speed?. I believe it is 0, however, I have been corrected that it is the combination of both, ie: 250 km.

Or was the above question incomplete?. Would rate of separation be the same as passing speed?.

If someone could explain to me and illustrate answer with formula, I would be very helpful. Thank you in advance.
 
Physics news on Phys.org
By passing speed I assume they mean the relative speed of one car with respect to the other. Just reason it out. If you were at rest and a car came towards you at 150 km/s, the relative speed would be 150 km/s. But if the other car were at rest and you were going towards it at a speed of 100 km/s, the relative speed would be 100 km/s. If you both move towards each other, add both speeds. Make some sense?

(And yes, it's the same as the rate of separation.)
 
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?

Similar threads

Replies
10
Views
2K
Replies
15
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
6
Views
2K
  • · Replies 51 ·
2
Replies
51
Views
5K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
6
Views
2K
Replies
1
Views
811
Replies
6
Views
2K
Replies
4
Views
2K