- #1
AidenPhysica
- 20
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Homework Statement
A plane flies 413 km east from city A to city B in 49.0 min and then 814 km south from city B to city C in 1.70 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? Give your angles as positive or negative values of magnitude less than 180 degrees, measured from the +x direction (east).
Homework Equations
so velocity is = to change in displacement over change in time
The Attempt at a Solution
For a, the magnitude of displacement is (413km^2+814^2)^.5 pythagorean theorem. is 912.8 km right?
For b, the direction of the plane's displacement is tan^-1 (814.2/413). but this angle 63.104 must be expressed of magnitude less than 180 degrees measured from +x direction (east). So how would you do that? Because 63.104 degrees is positive and not right, do you just minus 180 degrees? Then is it just -116.902 degrees? Yeah I guess so.
For c, the magnitude of average velocity is displacement 912.8km/ 2.52 hours to get 362.2222 km/hour.
For d, is the direction of avg velocity just the same as the answer to b, is it just -116.902 degrees also?
For e, isn't avg. speed just 486.9 km/hour?
Basically I have got answers but am really unsure if I am doing it right or wrong. For all I know, I am completely missing the point. Thanks again.