# A plane flies 457 km east from city A to city B

## Homework Statement

A plane flies 457 km east from city A to city B in 41.0 min and then 949 km south from city B to city C in 1.30 h. For the total trip, what are the (a) magnitude (in km) and (b) direction of the plane's displacement, the (c) magnitude (in km/h) and (d) direction of its average velocity, and (e) its average speed (in km/h)? Give your angles as positive or negative values of magnitude less than 180 degrees, measured from the +x direction (east).

## The Attempt at a Solution

Part A

457cos(0) + 949sin(270)

= 457i + -949j
lRl=sqrt[(457)^2+(-949)^2]
R=1053.3 km

Part B

arctan(-949/457) = -64.29 degrees

I am not sure if either of those are correct and as for the rest of the problem I need help on how to solve for those things.

Never mind everyone I finally got it

(c) Average velocity = Total displacement/total time
= 1053 km/(41+78)/60 = 530.9 km/h
(d) Direction is the direction of displacement; - 64.3 degrees

(e) Average speed = total distance/total time
= (457 + 949) km/1.9833 h = 709 km/h

You have the distance traveled and the time for each direction can't you calculate the velocity of each vector?

yea I got it now thanks for your help