Clock on an escalator

[Genady]

I've received this "riddle" from a friend this morning:

Doing my best to translate it to English here, while cutting off irrelevant details.

Bob got an electronic clock in shape of an apple. It shows time with a precision of a hundredth of a second. As he was moving down on an escalator, Bob threw the clock up and noticed that at the top of its trajectory the clock showed 11:32:45:81. His teacher Mary was moving up on the escalator at the same time, and she noticed that the clock showed 11:32:45:74 at the top of its trajectory. Find the speed of the escalators, given that they move with the same speed, at the angle of 30⁰ to horizon. Ignore the air friction. Take g = 10 m/s².

[Genady]

Hint: This problem is about comparing frames of reference.

[joigus]

Spoiler

The horizontal motion is just a distraction when it comes to calculating the respective times. I look at it as Bob being at rest and Mary going upwards at speed 2Vsin(30^o).

t_Bob=v_0z/g

t_Mary=v'_0z/g

where v_0z and v'_0z are the initial conditions for the vertical components of the velocity as measured by (resp.) Bob and Mary.

The transformation of velocities is:

v'_0z=v_0z-2V sin(30^o)=v_0z-V

where V is the total speed of both escalators. Naturally, Mary sees things going up "more slowly".

So that,

t_Bob-t_Mary=(v_0z-v'_0z)/g=V/g ==> V= (t_Bob-t_Mary)g=(10 ms^-2)(0.07 s)=0.7 ms^-1

 

[Genady]

6 minutes ago, joigus said:

  Hide contents

The horizontal motion is just a distraction when it comes to calculating the respective times. I look at it as Bob being at rest and Mary going upwards at speed 2Vsin(30^o).

t_Bob=v_0z/g

t_Mary=v'_0z/g

where v_0z and v'_0z are the initial conditions for the vertical components of the velocity as measured by (resp.) Bob and Mary.

The transformation of velocities is:

v'_0z=v_0z-2V sin(30^o)=v_0z-V

where V is the total speed of both escalators. Naturally, Mary sees things going up "more slowly".

So that,

t_Bob-t_Mary=(v_0z-v'_0z)/g=V/g ==> V= (t_Bob-t_Mary)g=(10 ms^-2)(0.07 s)=0.7 ms^-1

 

My calculations were a bit different, but the last line and the answer are the same. +1

Here is my reply to the friend who sent it to me:

Spoiler

Let's say the escalator moves up with a vertical speed v. Then Bob moves (down) with the vertical speed -v, and Mary moves up with the vertical speed v. Let's call the vertical speed of the clock relative to the ground at time t, u(t). The vertical speed of the clock relative to Mary is then u(t)-v, and its vertical speed relative to Bob is u(t)+v.

 

When the clock is in the highest point in Bob's reference frame, it stops momentarily, in this frame. Thus, at time tb=11:32:45:81,

u(tb)+v=0. 

 

In Mary's frame, it stops momentarily at the time tm=11:32:45:74. Thus, 

u(tm)-v=0.

 

Subtract the second equation from the first to get u(tb)-u(tm)=-2v. 

 

With the constant acceleration g=10 m/s^2, u(t)=-gt+const. Substitute it in the equation above to get g(tb-tm)=2v. Plug in the numbers and get v=0.35 m/s.

 

The escalator is moving at 30 degrees, so its linear speed is twice its vertical speed, which makes it 0.7 m/s.

 

[joigus]

17 minutes ago, Genady said:

My calculations were a bit different, but the last line and the answer are the same. +1

Here is my reply to the friend who sent it to me:

  Hide contents

Let's say the escalator moves up with a vertical speed v. Then Bob moves (down) with the vertical speed -v, and Mary moves up with the vertical speed v. Let's call the vertical speed of the clock relative to the ground at time t, u(t). The vertical speed of the clock relative to Mary is then u(t)-v, and its vertical speed relative to Bob is u(t)+v.

 

When the clock is in the highest point in Bob's reference frame, it stops momentarily, in this frame. Thus, at time tb=11:32:45:81,

u(tb)+v=0. 

 

In Mary's frame, it stops momentarily at the time tm=11:32:45:74. Thus, 

u(tm)-v=0.

 

Subtract the second equation from the first to get u(tb)-u(tm)=-2v. 

 

With the constant acceleration g=10 m/s^2, u(t)=-gt+const. Substitute it in the equation above to get g(tb-tm)=2v. Plug in the numbers and get v=0.35 m/s.

 

The escalator is moving at 30 degrees, so its linear speed is twice its vertical speed, which makes it 0.7 m/s.

 

Interesting...