total distance: 600+420=1020 km
total time: 1 day + 2hr = 26 hr
avg speed: 1020km/26hr = 39.23 km/hr
total displacement = √(600^2+420^2) = 732.39 km
in a direction θ where tanθ = 420/600, so θ = 35°
So, avg velocity is 732.39/26 = 28.17 km/hr at N35°E
A boat start sailing on Monday 10:00 at harbour A and sails 600km due north. Later it turns east for a further 420km to reach harbour B on Tuesday at 12:00..calculate the average velocity of the trip
1 answer
2 answers