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 begins a series to illustrate this based on the website *Physics: Problems and Solutions, Kinematics*.

Problem 3

A car travels up a hill at a constant speed of 37 km/h and returns down the hill at a constant speed of 66 km/h. Calculate the average speed for the whole trip.

Solution

Take the harmonic mean of the two speeds: 2/((1/37)+(1/66)) = 47.4 km/hr.

Problem 3.1

A car travels up a hill at a constant pace of 97.3 s/km and returns down the hill at a constant pace of 54.5 s/km. Calculate the average pace for the whole trip.

Solution

Take the average of the two paces: 75.9 s/km ≡ 47.4 km/hr.

Problem 9

An airliner reaches its takeoff speed of 163 mph in 36.2 s. What is the magnitude of its average acceleration?

Solution

By definition the acceleration is the ratio of the change of velocity and the traveled time. In the present problem the change of velocity is 163 mph = 0.447 × 163 m/s = 72.9 m/s. The traveled time is 36.2 s. Then the average acceleration is 72.9 / 36.2 = 2 m/s².

Problem 9.1

An airliner reaches its takeoff pace of 22.1 s/mi (≡ 163 mph) after traveling 100 yds down the runway. What is the magnitude of its average expedience?

Solution

By definition the expedience is the ratio of the change of legerity and the distance traveled. In the present problem the change of legerity is 22.1 s/mi × 1.60934 = 13.7 s/km. The distance traveled is 100 yds × 0.0009144 = 0.09 km. Then the average expedience is 13.7 / 0.09 = 152.2 s/km².

Problem 18

An athlete swims the length of a 50-m pool in 20 s and makes the return trip to the starting position in 22 s. Determine their average speed in (a) the first half of the swim, (b) the second half of the swim, and (c) the round trip.

Solution

The average speed is the ratio of distance traveled distance and travel time. Then (a) 50/20 = 2.5 m/s, (b) 50/22 = 2.27 m/s, and (c) (50+50)/(20+22) = 2.38 m/s.

Problem 18.1

An athlete swims the length of a 50-m pool in 20 s and makes the return trip to the starting position in 22 s. Determine their average pace in (a) the first half of the swim, (b) the second half of the swim, and (c) the round trip.

Solution:

The average pace is the ratio of travel time and distance traveled distance. Then (a) 20/50 = 0.4 s/m, (b) 22/50 = 0.44 s/m, (c) (20+22)/(50+50) = 0.42 s/m.