- #1

AidenPhysica

- 20

- 0

## 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.