Galilean transformation problem (Speed)

AI Thread Summary
A girl riding a bicycle passes a friend at a bus stop, and 60 seconds later, the friend catches a bus, with a distance of 126 meters between the events in the cyclist's frame. The calculation for the bicycle's speed yields a negative value of -2.1 m/s, which raises questions about its significance since speed is a scalar. The negative result indicates the direction of motion relative to the reference frame, where the cyclist perceives the ground moving backward. In the cyclist's frame, the events occur at 0 and -126 m, confirming that the distance is measured from the cyclist's perspective. Understanding the reference frames is crucial for interpreting the results correctly.
Ace.
Messages
52
Reaction score
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?
 
Physics news on Phys.org
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
 
  • Like
Likes Juwonjay
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?
 
Ace. said:
Why wouldn't it be +126 m if she is moving forward?
To the cyclist, the ground is moving backwards.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top