Question

The rodeo clown zigzagged in the ring to avoid the stampeding hooves of an upset horse....

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

Homework Answers

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.

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