Finding the velocity of a car in a different frame of reference

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The discussion focuses on calculating the velocity of a car from different frames of reference, specifically comparing the road (frame S) and a second car (frame S'). The initial equation for the first car's velocity is given as v = v_0 + at. The Galilean transformation is introduced, leading to confusion about the correct expression for V, which represents the speed of the second car relative to the first. Clarification is provided that V is indeed the speed of the second car with respect to the road, and the final equation simplifies to v' = V, confirming the relationship between the two frames. The conversation highlights the need for precise wording in physics problems to avoid misunderstandings.
Redwaves
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Homework Statement
A car is moving on a road with a initial speed ##v = v_0## and then it starts to speed up with ##a_0## what's the speed of this car in the frame of reference on a second car moving at the speed V.
Relevant Equations
##v' = (v_0 + at) - V##
Here's what I did so far.
The velocity of the first car is ##v = v_0 +at##

Frame of reference S = the road
Frame of reference S' = the second car

thus, v' is the speed of the first car in the frame of reference S' and v the speed in the frame of reference S.

Here's what make me doubt.
The Galilean transformation
##v' = v - V##
V should be the speed between S and S', in this case what I wrote is wrong.
V should be ##(v_0 + at) - V ##, right?

And then, ##v' = (v_0 + at) - ((v_0 + at) - V)## does it make sense ?
 
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Redwaves said:
Here's what make me doubt.
The Galilean transformation
##v' = v - V##
V should be the speed between S and S', in this case what I wrote is wrong.
V should be ##(v_0 + at) - V ##, right?

And then, ##v' = (v_0 + at) - ((v_0 + at) - V)## does it make sense ?
You were correct the first time. V is the speed of the second car (and thus frame S') with respect to S.

Realize that your final equation becomes ##v' = (v_0 + at) - ((v_0 + at) - V) = V##. Does that make sense?
 
I see. I didn't realize that the road is at rest... Of course the speed between S and S' is V.
Thanks!
 
Redwaves said:
Homework Statement:: A car is moving on a road with a initial speed ##v = v_0## and then it starts to speed up with ##a_0## what's the speed of this car in the frame of reference on a second car moving at the speed V.
If someone is going to set questions like this, they ought to be more precise, IMO:

A car is moving on relative to a road with an initial speed velocity ##v = v_0## and then it starts to speed up with accelerate at ##a_0##. What's the speed velocity of this car in the frame of reference of a second car moving at the speed velocity V relative to the road.
 
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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