Galilean transformation problem (Speed)

In summary: So the displacement of the bus stop person appears to be in the negative direction. In the bsp's frame, the cyclist's displacement would be positive since she is moving forward.
  • #1
Ace.
52
0

Homework Statement



A girl is riding a bicycle along a straight road at constant speed, and passes a friend standing at a bus stop (event #1). At a time of 60 s later the friend catches a bus (event #2)
If the distance separating the events is 126 m in the frame of the girl on the bicycle, what is the bicycle's speed?

Homework Equations


u = u' + v




The Attempt at a Solution


u = u' + v
can be written as:
Δx/t = Δx'/t + v
v = Δx/t - Δx'/t
v = 0m/60s - 126m /60s
v = -126 m/ 60s
v = -2.1 m/s

Just wondering if the negative holds any significance? I know we're talking about speed which is scalar but how come the calculation gives a negative?
 
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  • #2
Ace. said:

Homework Statement



A girl is riding a bicycle along a straight road at constant speed, and passes a friend standing at a bus stop (event #1). At a time of 60 s later the friend catches a bus (event #2)
If the distance separating the events is 126 m in the frame of the girl on the bicycle, what is the bicycle's speed?

Homework Equations


u = u' + v




The Attempt at a Solution


u = u' + v
can be written as:
Δx/t = Δx'/t + v
v = Δx/t - Δx'/t
v = 0m/60s - 126m /60s
v = -126 m/ 60s
v = -2.1 m/s

Just wondering if the negative holds any significance? I know we're talking about speed which is scalar but how come the calculation gives a negative?
It should be +2.1 m/s. This is because the Δx' represents the displacement of the second event minus the displacement of the first. So v = 0/60 - (-126/60).

You are using the cyclist's reference frame with the origin at the cyclist to determine u' and the bus stop person's (bsp) reference frame with the bsp at the origin to determine u. There are two events: 1. the origins coincide and 2. the bsp enters the bus.

In the cyclist's frame these events occur at 0 and -126 m. using the direction of v as the +x direction. So Δx' = -126m. In the bsp's frame, they both occur at the origin.

AM
 
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  • #3
Andrew Mason said:
In the cyclist's frame these events occur at 0 and -126 m.

Why wouldn't it be +126 m if she is moving forward?
 
  • #4
Ace. said:
Why wouldn't it be +126 m if she is moving forward?
To the cyclist, the ground is moving backwards.
 
  • #5




The negative sign in this calculation indicates that the direction of the bicycle's motion is opposite to the direction of the bus. This is known as a relative velocity and is commonly used in Galilean transformations to describe the motion of objects in different reference frames. In this case, the negative sign does not affect the magnitude of the bicycle's speed, which is 2.1 m/s, but it does indicate the direction of its motion. It is important to pay attention to the signs in these types of calculations, as they can provide valuable information about the motion of objects in different reference frames.
 

1. What is the Galilean transformation problem?

The Galilean transformation problem is a mathematical challenge that arises from the principles of Galilean relativity. It involves transforming coordinates and velocities between different frames of reference.

2. How does the Galilean transformation solve the speed problem?

The Galilean transformation allows us to calculate the relative velocity between two frames of reference by simply adding or subtracting their velocities. This solved the "speed problem" by providing a consistent method for determining relative speeds between moving objects.

3. What are the limitations of the Galilean transformation?

The Galilean transformation is only valid for objects moving at speeds much less than the speed of light. It also does not take into account the effects of relativity, such as time dilation and length contraction.

4. How does the Galilean transformation relate to Newton's laws of motion?

The Galilean transformation is based on the principles of Galilean relativity, which are consistent with Newton's laws of motion. This means that the laws of motion are the same in all inertial frames of reference, regardless of their relative speeds.

5. Can the Galilean transformation be applied to non-inertial frames of reference?

No, the Galilean transformation is only applicable to inertial frames of reference where Newton's laws of motion hold true. For non-inertial frames, more complex transformations, such as the Lorentz transformation, must be used to account for the effects of acceleration and gravity.

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