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Unformatted text preview: 1-1) a) We observe that the book slows down once it leaves your hand. Newton's first law says that an object in motion will remain in motion unless a force acts upon it. The book is slowing down (changing the rate of motion), so we know a force is acting upon it. We know from experience that this force is friction between the book and the table. b) We assume that the force from air drag is very small and thus has a negligible effect. If the arrow is moving very fast, the acceleration caused by gravity will not be noticeable over a short period of time. Of course, if we followed the complete trajectory of the arrow it would eventually fall to the ground. Basically the effects of these forces will be so small that we just ignore them. c) The cheetah is moving very fast, so it must work very hard to keep its legs moving as fast as its motion relative to the ground. If the cheetah stopped moving its legs, they would be in constant contact with the ground. This would generate a lot of friction and would slow the cheetah down, which is why the cheetah must move its legs just to stay at the same speed. 1-2) 1-3) a) Using the same logic as in the text discussion, the velocity of car B with respect to ground must be equal to the sum of the velocity of car B with respect to car A and the velocity of car A with respect to ground. VA/g = VA/B + VB/g VA/B = VA/g - VB/g b) Just plugging into the above formula: VA/B = 101 km/h 88 km/h = +13 km/h c) Start with the Eq. 2 formula for VA/B and then realize that it is simply opposite in sign to the formula 9in (a) for VB/A. VB/A = VB/g - VA/g ! - (-VB/g + VA/g ) " VB/A = - VA/B d) To find VA/A, use Eq. 2 and replace B with A, so that VA/A = VA/g - VA/g = 0 ! VA/A = 0 Similarly, use the formula from part (a) and replace A with B to find VB/B: VB/B = VB/g - VB/g VB/B = 0 e) They are true for any A and B. frames A, B, g. Eq. 2, for example, refers to any three reference ...
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- Spring '08