Solving Physics Problems (Part 2)

Continuing with solving problems from Understanding Relativity by Sartori.

Problem 1.3:

A jetliner has an air speed of 500 mph.  A 200 mph wind is blowing from west to east.
(a) The plane heads due north.  What direction does the plane fly and what is its ground speed?
Answer:  First, let us define that the S frame shall be as observed on the ground and the S' frame shall be as observed in the plane.  The magnitude of the plane's velocity with respect to the surrounding air is always 500 mph.  Thus we can say that:

The Galilean transformation from S' to S is:

And the equations of motion in S are:

From this we can say the direction the plane flies in is arc tan (200/500) = 21.8 degrees east of north.  The ground speed is

(b) In what direction should the pilot head in order to fly due north?  What is the plane's ground speed in this case?
Answer:  We are trying to find the angle such that the x component of the velocity in the ground frame is 0.  Using the Galilean transformation, we determine that the x component of velocity in the plane's frame has to be -200 mph, but the magnitude still has to remain 500 mph.  So to get the y component, we need to solve:

The direction the plane flies in is arc tan (200/458.3) = 23.6 degrees west of north and the ground speed is 458.3 mph.

Problem 1.4

A river is 20 m wide; a 1 m/s current flows downstream. Two swimmers, A and B, arrange a race.  A is to swim to a point 20 m downstream and back while B straight across and back.  Each can swim 2 m/s in the water.

(a) in what direction should B head in order to swim straight across.  Illustrate with a sketch.

Answer: In the ground frame S, the total magnitude of B's velocity is 2 m/s.  And we know that the x-direction component is 1 m/s.  Using the Pythagorean theorem, we have:

Taking the arc tan of the two sides, we find that the heading is 30 degrees (π/6 rad) towards the upstream from the straight across position.  Here is the sketch:

(b) Who wins the race and by how much time?

Answer: This question is important since the book takes us through the Michelson-Morley ether detection experiment in the following section.

The speed of B, going straight across and back remains the same the entire time at square root 3 m/s and as such, the elapsed time is 23.09 seconds.

The speed of A changes between the downstream and upstream legs.  For going down stream, she goes at 3 m/s taking 6 2/3 seconds, but going upstream, she goes 1 m/s and takes 20 seconds for a total time of 26.7 seconds.  Thus B wins by about 3 1/2 seconds.