There is a tree in the middle of your front yard and you take pictures of a squi

Question

There is a tree in the middle of your front yard and you take pictures of a squirrel (with a mass of 0.50

kg) as it runs around the tree. When you take the rst picture, the squirrel 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 squirrel 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

squirrel 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 squirrel between the rst and second pictures?

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

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

(d) What is the change in the momentum of the squirrel 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, thenet displacementcan 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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