1. There is a tree in the middle of your front yard and you take pictures of a s

Question

1. There is a tree in the middle of your front yard and you take pictures of a squirrell as it runs around
the tree. When you take the rst picture, the squirrell is 12 m away from the tree in the direction 30°
N of E moving with a speed of 1.8 m/s due east. Twelve seconds later, you take a second picture and
the squirrell is 7.5 m away from the tree in the direction 15° E of S moving due north with a speed
of 1.5 m/s. After an additional 15 seconds, you take a third picture when the squirrell is 9.5 m away
from the tree in the direction 25° W of S moving with a speed of 2.5 m/s due west.

(a) What is the displacement of the squirrell between the rst and second pictures?

(b) What is the average velocity of the squirrell between the second and third pictures?

(c) What is the average acceleration of the squirrell between the rst and third pictures?

Explanation / Answer

Part A)

We can solve this by the law of cosines.

The magnitude of the two dispacement vertors that we know are 12 m and 7.5 m. Based on the angles given, it can be shown that the total angle between those two vecotrs is 95o

Therefore, the net displacement can be found by

c2 = a2 + b2 -2abcosC

c2 = (12)2 + (7.5)2 - (2)(12)(7.5)(cos95)

c = 14.7 m

Part B)

Between the second and third picture the magnitudes of the displacement vectors that we know ar 7.5 m and 9.5 m. The angle between them is 40o, so

c2 = a2 + b2 -2abcosC

c2 = (9.5)2 + (7.5)2 - (2)(9.5)(7.5)(cos40)

c = 6.11 m

Then since d = vt,

The velocity = d/t

v = (6.11)/(15)

v = .41 m/s

Part C)

The average acceleration can be found by...

vf = vo + at

-2.5 = 1.8 + a(27)

a = -.159 m/s2    

The negative only symbolizes direction, so you can eliminate it if you want and call it .159 m/s2

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