A plane flies 483 km east from city A to city B in 45.0 min and then 966 km south from city B to city C in 1.50 h. For the total trip, what are the (a) magnitude and (b) direction of the plane’s displacement, the (c) magnitude and (d) direction of its average velocity, and (e) its average speed?

a.1080km

b.63.4° S of E

c.480km/h

d.63.4° S of E

e.644km/h

Explanation:

a. Using Pythagorean theorem:

The planes magnitude is therefore 1080km

b. Given the plane's directions as 483km east and 966 west, the angle of displacement can be obtained using Trigonometric properties as:

The plane's displacement is therefore given as 63.4° S of E

c. From a, above we know the plane's magnitude as 1080km.

#We also know the time taken as 1.5h. By definition average velocity is distance over time:

             #D-displacement, t=0.75+1.5=2.25h

Hence, the average velocity of the plane is 480km/h

d. Given the plane's directions as 483km east and 966 west, the angle of displacement can be obtained using Trigonometric properties as:

The plane's displacement is therefore given as 63.4° S of E

e. Average speed is the total distance divided by total time taken.

Total time =45min+1.5h=2.25

Total distance=483+966=1449km

Therefore the speed is calculated as:

The average speed of the plane is 644km/h