**Problem : **
Someone on a moving train on the earth measures the speed of a meteor in
space to be 5×10^{6} m/s. Someone in outer space measures the
speed to be 4×10^{6} m/s. Who is right?

**Problem : **
Two spaceships are hurtling towards one another at a constant speed of
0.8*c*. When they are still 10 000 kilometers apart, one spaceship radios
the other to warn them of the impending collision. How much time does it
take for the radio wave to reach the other ship, as observed by someone on
the receiving ship (assume that the spaceships move little in the time
taken for the signal to travel between them)?

*c*, according to our second postulate. Thus the time taken is just

*t*=

*d*/

*v*= 10000/3×10

^{8}= 3.33×10

^{-5}m/s.

**Problem : **
Consider the situation described in Section 1. If the flashes
from the sources are observed to occur simultaneously by an observer
standing on the ground (at rest relative to the sources), what is the time
difference between the events according to an observer on a train speeding
past at 0.15*c*, if that observer measures the distance between the
sources to be 1 kilometer?

*l*= 500m. Then

*t*

_{r}= = = 1.96×10

^{-6}seconds, and

*t*

_{l}= = = 1.45×10

^{-6}seconds. Thus the time difference is

*t*

_{r}-

*t*

_{l}= 5.12×10

^{-7}seconds. Even at the immense speed of 45000 kilometers per second, the time difference is hardly noticeable.

**Problem : **
What if the scenario described in Section 1 is performed with
baseballs (which travel at a constant speed *b* < *c*) instead of light
pulses. Will the observers still disagree?

*O*

_{A}would still see the baseballs arrive simultaneously and conclude that they were thrown simultaneously.

*O*

_{B}sees the baseball on the source on the right thrown with speed

*b*-

*v*and the baseball from the source on the left thrown with speed

*b*+

*v*.

*O*

_{B}then calculates the speeds relative to the throwers as (

*b*-

*v*) +

*v*=

*v*on the left and (

*b*+

*v*) -

*v*=

*v*on the right. Thus

*O*

_{B}too concludes that the baseballs arrive simultaneously. There is no disagreement; this comes about as a consequence of the weird properties of

*c*.

**Problem : **
Consider again the scenario described in Section 1. Now
consider changing the setup by placing only a single emitter at the center
position (where *O*_{A} was), and having two receivers placed where the
sources formerly were. The source emits two signals, one in each
direction (that is, one towards each receiver). An observer at rest with
respect to the source and receivers concludes that the source emitted its
two signals simultaneously. What does an observer traveling to the right
at velocity *v* observe?

*t*

_{l}= to reach the left receiver, and the right-moving light takes a time

*t*

_{r}= to reach the right receiver. Thus the 'left' and 'right' times are swapped.