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exam1practiceSolutions

# exam1practiceSolutions - Updated for 2007 1 Convert all of...

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Updated for 2007!!! 1. Convert all of the following into proper mks units: (a) 2500 feet hr OK, we have two problems with the above expression, feet should be meters and hr should be seconds. We know that 1 foot = 12 inches, 1 inch = 2.45 cm, and 1 m = 100 cm. So: 2500 feet hr × 12 inches 1 foot × 2 . 54 cm 1 in × 1 m 100 cm × 1 hr 60 min × 1 min 60 sec (1) 2500 × 12 × 2 . 54 100 × 60 × 60 m s (2) 0 . 212 m s (3) (b) 3 miles 3 miles × 5280 feet 1 mile × 12 inches 1 foot × 2 . 54 cm 1 in × 1 m 100 cm (4) 3 × 5280 × 12 × 2 . 54 100 meters (5) 4828 meters (6) (c) 2 years 2 years × 365 days 1 year × 24 hours 1 day × 60 minutes 1 hr × 60 sec 1 min (7) 2 × 365 × 24 × 60 × 60 1 sec (8) 6 . 31 × 10 7 sec (9) (d) 12000 cm 3 12000 cm 3 × 1 m 100 cm × 1 m 100 cm × 1 m 100 cm (10) 12000 100 × 100 × 100 m 3 (11) 0 . 012 m 3 (12) (e) 45 gram × cm minute 2 1

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45 gram × cm minute 2 × 1 kg 1000 grams × 1 m 100 cm × 1 min 60 sec × 1 min 60 sec (13) 45 1000 × 100 × 60 × 60 kg × m s 2 (14) 1 . 25 × 10 - 7 kg × m s 2 (15) 1 . 25 × 10 - 7 N (16) 2

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40 = 30 t - 5 t 2 (28) 5 t 2 - 30 t + 40 = 0 (29) t = - b ± b 2 - 4 ac 2 a (30) t = 30 ± 30 2 - 4 × 5 × 40 2 × 5 (31) t = 30 ± 900 - 800 10 (32) t = 30 ± 100 10 (33) t = 30 ± 10 10 (34) t = 3 ± 1 (35) Since we clearly need the earliest time as the time you hit the deer, we get: t = 2 sec (36) Luckily this is one smart deer. The deer runs to the side and needs to move a minimum of 1.2 meters to avoid being hit. What acceleration of the deer is necessary to just avoid the collision with your car? ANSWER: The deer must achieve the position of 1.2 meters when the car arrives (t = 2 sec). This “definition of success is written as: x deer ( t = 2) = 1 . 2 meters This is solved using the x-t relation (while being very careful with nota- tion!): x deer = x 0 deer + v 0 deer t + 1 2 a deer t 2 (37) 1 . 2 = 0 + 0 + 1 2 a deer 2 2 (38) 1 . 2 = 2 × a deer (39) a deer = 0 . 6 m s 2 (40) 4
3. Two trains are on the same track. Train A is in the back and is moving with a constant velocity, v 0 A = 12 m s . Train B is ahead of train A by 50 m and is moving with a velocity of v 0 B = 10 m s . Just at this moment Train B spots the trouble and begins to accelerate.

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