Answer:
40.3 min
Explanation:
First of all, let's convert every quantity into SI units:
[tex]v_1 = 92.5 km/h = 25.7 m/s[/tex] (speed in the first part of the trip)
[tex]t_2 = 28.0 min = 1680 s[/tex] time during which the person has stopped
[tex]v=72.2 km/h = 20.1 m/s[/tex] (average speed of the whole trip)
The average speed is the ratio between the total distance covered, d, and the total time taken, t:
[tex]v=\frac{d}{t}[/tex] (1)
The total distance covered is simply
[tex]d = v_1 t_1[/tex]
where [tex]t_1[/tex] is the time during which the person has moved at 92.5 km/h.
The total time taken is
[tex]t= t_1 + t_2[/tex]
So (1) becomes
[tex]v=\frac{v_1 t_1}{t_1 + t_2}[/tex]
Solving for [tex]t_1[/tex]:
[tex]v t_1 + v t_2 = v_1 t_1\\vt_2 = (v_1+v)t_1\\t_1 = \frac{v t_2}{v_1+v}=\frac{(20.1 m/s)(1680 s)}{25.7 m/s + 20.1 m/s}=737.3 s[/tex]
which corresponds to
[tex]t_2 = 737.3 s = 12.3 min[/tex]
So the total time of the trip is
[tex]t = 28.0 min + 12.3 min = 40.3 min[/tex]
