Complete Question:
Each of the following trips lasts exactly one hour. The positive direction is to the east.
(i) Automobile A travels 50 km due east. (ii) Automobile B travels 50 km due west. (iii) Automobile C travels 60 km due east, then turns around and travels 10 km due west. (iv) Automobile D travels 70 km due east. (v) Automobile E travels 20 km due west, then turns around and travels 20 km due east. (a) Rank the five trips in order of average x-velocity from most positive to most negative. (b) Which trips, if any, have the same average x-velocity? (c) For which trip, if any, is the average x-velocity equal to zero?
Answer:
a) 1. (iv) 2.) (i) and (iii). 3. (v) 4.(ii)
Explanation:
By definition, the average velocity, can be expressed as the total displacement (the difference between the final and initial position), divided by the time interval during which we define the displacement.
We can put these words in an equation format as follows:
[tex]vavg,x = \frac{xf-xo}{t-to}[/tex]
If we choose x₀ = 0 and t₀ = 0, the average velocity is just the final position over one hour.
The final positions, in the five trips, are as follows:
(i) = +50 Km (2nd)
(ii) = -50 Km (4th)
(iii) + 60 Km -10 Km = + 50 km (2nd)
(iv) = + 70 km (1st)
(v) = -20 km + 20 Km = 0 (3rd)
b) As stated above, the trips (i) and (iii) have the same average x-velocity.
c) As stated above, in trip (v) as the total displacement is 0, average -velocity is 0 also.