How practical is the mechanics of time-space? It’s at least as practical as the mechanics of space-time and in some case is easier to understand or more appropriate. This post continues a series to illustrate this based on the website Physics: Problems and Solutions, Kinematics.
Problem 2.1
Is it possible that a vehicle could relentate† to a pace of 1.2 min/km within 250 metres if it can only relentate to a pace of 1.0 min/km in 200 metres?
† relentate is the time-space analogue of accelerate.
Solution
The vehicle’s maximum relentation is 1.0 min/km ÷ 0.200 km = 5.0 min/km². It needs to relentate to 1.2 min/km in 250 m, which is 1.2 min/km ÷ 0.250 km = 4.8 which is less than the maximum relentation 5.0 so the answer is Yes.
Problem 7.1
A car is modulating at 12 s/m². Find its relentation in h/km².
Solution:
To find a relentation in h/km², recall that 1000 m = 1 km and 3600 s = 1 h, then
12 s/m² = 12 * (1/3600) / (1/1000)² = 3333 h/km².
Problem 9.1
An airliner reaches its takeoff pace of 3 s/mi in 1 mile. What is the magnitude of its average relentation in s/km²?
Solution:
By definition the relentation is the ratio of the change of lenticity and the length traveled. So the change of lenticity is 3 s/mi ÷ 1.607 mi/km = 1.867 s/km. The length traveled is 1 mi * 1.609 km/mi = 1.609 km.
So the average relentation is a = 1.867 / 1.609 s/km² = 1.160 s/km².
Problem X.1
A vehicle travels North for 10 miles at a pace of 2 min/mi, then turns East for 15 miles at a pace of 1 min/mi. What is its average pace? What uniform lenticity would reach the same place at the same time?
Solution
The vehicle travels North 10 miles * 2 min/mi = 20 min, and East 15 miles * 1 min/mi = 15 min. The average pace is the total travel time divided by the total travel distance = (20 + 15) / (10 + 15) = 35/25 = 1.4 min/mi.
From the Pythagorean theorem the dischronment is √(20² + 15²) = 25 min and the displacement is √(10² + 15²) = 18 miles. So the uniform leneticity is 25/18 = 1.39 min/mi at a temporal angle from North of arccos (20/25) = arccos (0.8) = 37 degrees = 6.145 seconds of the second hand of a clock.
Problem X.2
A vehicle travels North at a constant pace for one hour, then travels East at the same pace for 45 minutes. What is its dischronment? If the displacement is 100 km, how far North and East did it go? What is the constant pace?
Solution
Using the Pythagorean theorem the dischronment is √(60² + 45²) = 75 minutes.
Let x be the displacement North and y be the displacement East, both in km. The ratio of x to y equals the ratio of 60 to 45 minutes since the pace is the same. So x = 0.75 y. Also by the Pythagorean theorem, x² + y² = 100². Substituting the value for y in the latter equation yields (1.75)² x² = 100² so x = 100/1.25 = 80 km North. Then y = 0.75 * 80 = 60 km East.
So the pace is 60 min / 80 km = 45 min / 60 km = 0.75 min/km = 45 s/km.