I've started Unit 2 motion by myself and I'm not getting the answer for this question in the textbook. Part b. of the attached image,it looks rather straight forward but I'm not entirely sure what the first step is.
(Sorry about posting this here, but I think it's best to post it here since motion overlaps)
Here is what I think:
For part (a), a quick prediction would be 90 km/h, since it is half way between 80 km/h (from Melbourne to Wodonga) and 100 km/h (back from Wodonga to Melbourne).
For part (b), find the total distance and total time covered throughout the journey.
Total Distance = 300 km + 300 km = 600 km
Now speed = distance / time, so time = distance / speed
Time covered between Melbourne to Wodonga = (300 km) / (80 km/h) = 3.75 hours
Time covered between Wodonga to Melbourne = (300 km) / (100 km/h) = 3.00 hours
Therefore, total time = 3.75 hours + 3.00 hours = 6.75 hours
Hence, average speed = total distance / total time = 600 km / 6.75 h = 89 km/h (to 2 significant figures)
Reason for difference in predicted and calculated speeds:
90 km/h would be correct if we were considering a case where the initial speed was 80 km/h and the final speed was 100 km/h (i.e. using the formula v (av) = [u + v] / 2.
However, this is not the case. We are told the average speed for one interval of the journey, and the average speed for another interval. So the formula v (av) = Δ x / Δ t must be used, which yields 89 km/h.
I hope this helps!