# Thoughts Experiment about Frame of References

• kasha

## Homework Statement

The Observer (me) is at Inertial Frame of Reference:

1)
I am sitting at a car moving east v=30 km/h relative to earth
A bird flying east v = 10 km/h relative to earth
What is the speed of bird for me?

2)
I am sitting at a car moving east v=30 km/h relative to earth
A bird flying east v = 40 km/h relative to earth
What is the speed of bird for me?

3)
I am sitting at a car moving east v=30 km/h relative to earth
A bird flying east accelerating at a=10km/h2 relative to earth
What is the acceleration of the bird relative to me?

4)
I am sitting at a car moving east v=30 km/h relative to earth
A bird flying east accelerating at a=40 km/h2 relative to earth
What is the acceleration of the bird relative to me?

Please feel free to correct any stupidity/misconceptions I might have introduced in my questions

## Homework Equations

Frames of Reference

## The Attempt at a Solution

1) I think (-20km/h)...if the car and the bird started at the same spot, after 1hr, I would have traveled 30 km to the east, while the bird would have traveled only 10km to the east, so the bird is lagging behind me by 20km..looking as the bird is moving 20km/h to the west!

2) I think (+10km/h)...if the car and the bird started at the same spot, after 1hr, I would have traveled 30 km to the east, while the bird would have traveled 40km to the east, so the bird is a head of me by 10 km..looking as the bird is traveling 10km/h to the east!

3) 10 km/h2 ?!
4) 40 km/h2 ?!

Looks good to me.

Looks good to me.
Thanks but I was looking for more explanation.

for question (1 & 2). we simply use Galilean Transformation (let's not worry about Lorentz transformation)
The Galilean transformation gives the coordinates of the point as measured from the fixed frame in terms of its location in the moving reference frame.
The fixed frame is the Earth, the point moving is the bird, the frame moving (car)
but for the observer inside the car (the fixed frame is the car, the moving frame is the Earth moving at -30 km/h, the point moving is the bird)

x' (position of bird in moving frame "Earth") = x (position of bird measured from my frame "car") - vt (velocity of moving frame "Earth" with respect to mine)

for velocity transformation, we differentiate by t:
d(x')/dt = d(x)/dt - d(vt)/dt
v' (velocity of bird relative to Earth) = v (velocity of bird relative to car) - vs (velocity of car relative to Earth which is -30)

1)
10 = v - (-30)
v = -20 km/h (velocity of bird relative to car)

2)
40 = v - (-30)
v = 10 km/h (velocity of bird relative to car)

for (3&4) the acceleration transformation of the Galilean will be:
a' (acceleration of bird relative to Earth) = a (acceleration of bird relative to me)