Question

A bird flies at 12.5m/s due north from a tree, then at 20.0m/s for 3.0 minutes at 20 degrees south of west.

(I) If the bird's net displacement is 4.5km, 48.73 degrees west of north, what is the distance traveled and how much time has elapsed when the bird flies due north?

(II) What is the bird's average speeding during the entire journey?

Thanks

Answer 1

i) let East be +x axis

R = 4.5 km

Rx = -4.5*cos(48.73) = -2.968 km
Ry = 4.5*sin(48.73) = 3.382 km

let A is the distance tarvelled towards north

B is the distance travelled along 20 degrees south of west

Rx = Ax + Bx

Rx = 0 + Bx

Rx = Bx

Ry = Ay + By

3382 = Ay - 20*(3*60)*sin(20)

Ay = 3382 + 20*3*60*sin(20)

= 4613 m or 4.613 km <<<<<<<<<----------Answer

time taken, t = 4613/12.5

= 369 s or 6.15 minutes <<<<<<<<<----------Answer

ii)

Bird's average speed = total distance travelled/total time taken

= (4613 + 20*3*60)/(369 + 3*60)

= 14.96 m/s <<<<<<<<<-----------------Answer