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Kaplunovsky - practice 2

# Kaplunovsky - practice 2 - practice 02 Due noon Question 1...

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practice 02 – – Due: Jan 29 2007, noon 1 Question 1, chap 2, sect 5. part 1 of 1 10 points A sailboat is initially moving at a speed of 2 m / s. A strong wind blows up and ac- celerates the boat forward with a constant acceleration of 0 . 1 m / s 2 for 10 s. What is the final speed of the sail boat? Correct answer: 3 m / s (tolerance ± 1 %). Explanation: The speed of a boat with initial speed v 0 with a constant acceleration a is given by v = v 0 + at Therefore the speed of the boat after time 10 s s is v = 2 m / s + (10 s) (0 . 1 m / s 2 ) = 3 m / s Question 2, chap 2, sect 6. part 1 of 1 10 points An object is released from rest on a planet that has no atmosphere. The object falls freely for 3 m in the first second. What is the magnitude of the acceleration due to gravity on the planet? 1. 6 . 0 m / s 2 correct 2. 1 . 5 m / s 2 3. 3 . 0 m / s 2 4. 10 . 0 m / s 2 5. 12 . 0 m / s 2 Explanation: Let : s = 3 m . s = 1 2 at 2 a = 2 s t 2 = 2 (3 m) (1 . 0 s) 2 = 6 m / s 2 . Question 3, chap 2, sect 5. part 1 of 2 5 points A motorist is traveling at 15 m / s when he sees a deer in the road 39 m ahead. If the maximum negative acceleration of the vehicle is - 7 m / s 2 , what is the maximum reaction time Δ t of the motorist that will allow him to avoid hitting the deer? Correct answer: 1 . 52857 s (tolerance ± 1 %). Explanation: Let : v 0 = 15 m / s , v = 0 , and a = - 7 m / s 2 . Basic Concept: v = x t v 2 = v 2 0 + 2 ax. Solution: v 2 = v 2 0 + 2 ax stop x stop = - v 2 0 2 a = 16 . 0714 m . x - x stop = v 0 Δ t. This yields Δ t = x - x stop v 0 = 1 . 52857 s . Question 4, chap 2, sect 5. part 2 of 2 5 points If his reaction time is 1 . 71747 s, how fast will he be traveling when he reaches the deer? Correct answer: 6 . 29832 m / s (tolerance ± 1 %). Explanation:

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practice 02 – – Due: Jan 29 2007, noon 2 Let: Δ t = 1 . 71747 s Solution: x slow = x - v 0 Δ t. We again use the relation v 2 = v 2 0 + 2 ax slow , which gives v = radicalBig v 2 0 + 2 a ( x - v 0 Δ t ) = 6 . 29832 m / s . Question 5, chap 2, sect 5. part 1 of 1 10 points Hint: x 1 and x 2 never have the same value. The x -coordinates of two objects moving along the x -axis are given below as a function of time t . x 1 = (4 m / s) t and x 2 = - (82 m) + (16 m / s) t - (1 m / s 2 ) t 2 . Calculate the magnitude of the distance of closest approach of the two objects. 1. 23 m 2. 46 m correct 3. 33 m 4. 39 m 5. 47 m 6. 54 m 7. 25 m 8. 36 m 9. 29 m 10. 51 m Explanation: Given : v 1 = 4 m / s , x 0 = 82 m , v 2 = 16 m / s , and a 2 = 1 m / s 2 . x 1 = v 1 t and x 2 = - x 0 + v 2 t - a 2 t 2 . Δ x = x 2 - x 1 = - x 0 + [ v 2 - v 1 ] t - a 2 t 2 . The distance of closest approach will be when the derivative d Δ x dt = v 2 - v 1 - 2 a 2 t = 0 . Therefore, t = v 2 - v 1 2 a 2 = (16 m / s) - (4 m / s) 2 (1 m / s 2 ) = 6 s . At that time, we have x 2 = - x 0 + v 2 t - a 2 t 2 = - (82 m) + (16 m / s) (6 s) - (1 m / s 2 ) (6 s) 2 = - 22 m x 1 = at = (4 m / s) (6 s) = 24 m Δ x = x 2 - x 1 = ( - 22 m) - (24 m) = - 46 m | Δ x | = 46 m .
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