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**Unformatted text preview: **Homework Assignment 3 Solutions Problem 2.33: Air-bag injuries. During an auto accident, the vehicle's air bags deploy and slow down the passengers more gently than if they had hit the windshield or steering wheel. According to safety standards, the bags produce a maximum acceleration of 60 g, but lasting for only 36 ms (or less). How far (in meters) does a person travel in coming to a complete stop in 36 ms at a constant acceleration of 60g? Answer: 38.1 cm Before we try to solve this problem, convert 36 ms into seconds: 36 ms 1 sec 1000 ms = 0.036 sec Then we need to use two kinematic equations. First we use v = v o at to find the initial velocity v o (or velocity before the braking started). We assume v = 0 (because it stops): v o = at v o = 60 9.8 m / s 2 0.036 s v o = 21.2 m / s Note that the acceleration is negative (i.e., a deceleration). Then, we use the second kinematic equation, v 2 = v o 2 2a x , to find the distance x traveled in 0.036 sec (remember v = 0 because it stops): x = v o 2 2a x = 21.2 m / s 2 2 60 9.8 m / s 2 x = 0.381 m x = 38.1 cm Problem 2.38 The "reaction time" of the average automobile driver is about 0.700 s. (The reaction time is the interval between the perception of a signal to stop and the application of the brakes.) If an automobile can slow down with an acceleration of 12.0 ft / s 2 , compute the total distance covered in coming to a stop after a signal is observed (a) from an initial velocity of 15.0 mi/h (in a school zone) and (b) from an initial velocity of 55.0 mi/h. A) Answer: 10.8 m First, convert 12.0 ft / s 2 into m/s: 12 ft 1 s 2 1 meter 3.28 ft = 3.7 m / s 2 And convert 15.0 mi/h into m/s: 15.0 mi 1 hr 1609 meters 1 mi 1 hr 3600 sec = 6.7 m / s Then to solve, we need to break the problem into two parts. The first part is when the car is not braking due to the fact that the driver has a reaction time. This distance d 1 , is easy to calculate: d 1 = v o t d 1 = 6.7 m / s 0.7 s d 1 = 4.69 m Then we need to use the kinematic equation v 2 = v o 2 2a x to find the second distance d 2 . In this equation, x is our d 2 and v = . d 2 = v o 2 2a d 2 = 6.7m / s 2 2 3.7 m / s 2 d 2 = 6.1 m Finally, we get d total by adding up d 1 and d 2 : d total = d 1 d 2 d total = 10.8 m B) 99.9 m First convert 55.0 m/h into m/s: 55.0 mi 1 hr 1609 meters 1 mi 1 hr 3600 sec = 24.6 m / s Then we solve it the same way that we solved A: d 1 = v o t d 1 = 24.6 m / s 0.7 s d 1 = 17.2 m d 2 = v o 2 2a d 2 = 24.6 m / s 2 2 3.7 m / s 2 d 2 = 81.8 m d total = d 1 d 2 d total = 99.9 m Problem 2.73 It has been suggested, and not facetiously, that life might have originated on Mars and been carried to ...

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