Answer:
a) Please find attached the required travel graph with times increasing from (0:00) of the day of travel to the following day
b) The average speed of return journey = 880 km/(15.5 h) ≈ 56.77 km/h
Explanation:
The given information includes;
The time of departure of the salesman, t₁ = 0800
The duration of travel of the salesman = 13 hours
The average speed with which the salesman travels = 60 km/h
The time duration in which he stops = 30 minutes
The speed with which he continues = 50 km/h
The time duration he travels at 50 km/h = 2 hours
The time he arrives home after he turns, t₂ = 16:00
The appropriate graph is a distance time graph
So as to show the given information in a travel graph, we calculate the distances as follows;
Time
Location
08:00
0
08:00 + 13 = 21:00
13 × 60 = 780
21:00 + 0.30 = 21:30
780
21:30 + 2:00 = 23:30
780 + 50 × 2 = 880
23:30 + 1:00 = 0:30 (24: hours )
880
16:00
(40 hours from 0:00 the previous day)
Therefore, we have;
Location; 0, 780, 780, 880, 880, 0
Time; 08:00, 21:00, 21:30, 23:30, 24:30, 40:00
To show the distance traveled over the given time periods graphically, we let 00:30 = 24.30 and the arrival time (the next day) 16:00 = 40.00 hours after midnight (0:00) the previous day
Please find attached the required travel graph created with Microsoft Excel
b) The average speed of return journey = (Total distance of return journey)/(Total time)
The total distance going = 880 km
Therefore the return journey is also 880 km
∴ The total time taken on return journey Δt = (Time at arrival home) - (Time of start of return journey)
∴ Δt = 40 - 24.5 = 15.5
Δt = 15.5 hour
∴ The average speed of return journey = 880 km/(15.5 h) ≈ 56.77 km/h.