Question

The rodeo clown zigzagged in the ring to avoid the stampeding hooves of an upset horse. She started by running West for 23 meters, then 12° East of South for 5 meters. Turning again, she ran 7 meters Northeast (45°). What was her displacement vector’s magnitude and direction?

Round intermediate answers to 4 significant figures and the final answer to 3 sig figs

Answer 1

First we take every displacement which are in different direction as different vectors.

First displacement:

West for 23 m, west is 180° from the east, which we are referring as positive x axis.

r1 = (23*Cos 180)*i + (23*Sin 180)*j

r1 = -23*i

Second displacement:

12° East of south for 5 m, total angle from x-axis = 270+12 = 282°

r2 = (5 * Cos 282)*i + (5*Sin 282)*j

r2 = 1.0395*i - 4.8907*j

Third displacement:

7 m Northeast, angle = 45°

r3 = (7*Cos 45)*i + (7*Sin 45)*j

r3 = 4.9497*i + 4.9497*j

Now the total displacement is,

r = r1 + r2 + r3

r = -17.0107*i + 0.059*j

where r is the displacement vector.

Magnitude of displacement vector:

|r| = √[(-17.0109)² + (0.059)²]

|r| = 17.010 m

Direction of displacement vector:

ß = arctan(0.059/17.0109)

ß = - 0.198°

Thus the direction is 0.198° South of East.