i went for a walk one day i walked north 6km at 6km/hr and then west 10km at 5km/hr [this problem is deceptively easy so be careful begin each part byreviewing the appropriate physical definition ] determine a the total distance of the entire trip b the total displacement of the entry trip c the average speed of the entry trip
To solve this problem, we will first analyze the distances traveled in each direction and then calculate the total distance, displacement, and average speed of the entire trip.
a) Total distance of the entire trip:
Distance traveled north = 6km
Distance traveled west = 10km
Total distance = 6km + 10km = 16km
b) Total displacement of the entire trip:
To calculate the displacement, we need to use the Pythagorean theorem as the displacement is the straight-line distance between the starting point and the ending point.
Displacement = √(6^2 + 10^2)
Displacement = √(36 + 100)
Displacement = √136
Displacement ≈ 11.66 km
c) Average speed of the entire trip:
Total time taken to travel north = Distance / Speed = 6km / 6km/hr = 1 hour
Total time taken to travel west = Distance / Speed = 10km / 5km/hr = 2 hours
Total time taken for the entire trip = 1 hour + 2 hours = 3 hours
Total distance traveled = 16km
Average speed = Total distance / Total time taken
Average speed = 16km / 3hr
Average speed ≈ 5.33 km/hr
Therefore, the answers to the questions are:
a) The total distance of the entire trip is 16 km
b) The total displacement of the entire trip is approximately 11.66 km
c) The average speed of the entire trip is approximately 5.33 km/hr.