Chapter_2_09

# Chapter_2_09 - Chapter 2 Problems 2.3 During a hard sneeze...

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Chapter 2 Problems 2.3 During a hard sneeze, your eyes might shut for 0.5s. If you are driving a car at 90km/h during such a sneeze, how far does the car move during that time s = 90 km h 1000 m 1 km 1 h 3600 s = 25 m / s d = s t = 25 m / s 0.5 s = 12.5 m / s 2.5 The position of an object moving in a straight line is given by x = 3 t 4 t 2 + t 3 , where x is in meters and t in seconds (a). What is the position of the object at t=1,2,3, and 4s? (b) What is the object’s displacement between t = 0 and t = 4s. (c) What is the average velocity for the time interval from t=2 s to t = 4s? (d) Graph x vs t for 0 t 4 s and indicate how the answer for c can be found from the graph. (a-d) We plug in to calculate positions. x (1) = 0 m x (2) = 2.0 m x (3) = 0 m x (4) = 12 m e) We can calculate the displacement from the positions. Δ x = x (4) x (0) = 12 m 0 m = 12 m (f) We calculate the average velocity using the displacements and time interval. v = x (4) x (2) 4 s 2 s = 12 m ( 2 m ) 2 s = 7 m / s (g) A graph of x vs t. . The average velocity can be computed by connecting x(4) and x(2) with a straight line and computing the slope. Note: Graph done with Mathematica.

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2.9 You are to drive to an interview in another town at a distance of 300 km on an expressway. The interview is at 11:15 AM. You plan to drive 100 km/h, so you eave at 8:00 AM to allow some extra time. You drive at that speed for the Frst 100 km, but then construction work forces you to slow to 40 km/h for 40 km. What would be the least speed needed for the rest of the trip to arrive in time for the interview? We need to compute how much time has passed and how far you have gone. If you drive at 100 km/h for 100 km, you spend 1 hour driving that segment. The next segment is 40 km at 40 km/ h. This takes 1 hour as well. So you have driven 2 hours and covered 140 km. You allowed 3.25 hours for the entire 300km trip and you now have 1.25 hours left to cover the remaining 160km. This means that you need to drive at v = 1.25 h 160 km = 128 h km to complete the trip on time. 2.11 Two trains, each having a speed of 30km/h, are headed at each other on the same straight track. A bird that can ±y 60km/h ±ies off the front of one train when they are 60 km apart and heads directly for the other train. On reaching the other train, the bird ±ies directly bact to the Frst train, and so forth. (We have no idea why a bird would behave in this way.) What is the total distance the bird travels before the trains collide? The easiest way to think about and do this problem is to note that the bird ±ies at a constant speed for the entire time that the trains travel until colliding. The distance the bird travels is thus d = v t =60 h km t We need to compute the time that the trains run before colliding. By symmetry, we can argue that the trains will collide in the middle, at 30 km. The time to collision is easy to compute--it’s
just the time for either time to travel 30 km, which is 1 hour. Now that we know how long the trains will run, we can see that the distance the bird will travel is 60 km.

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## This note was uploaded on 09/27/2010 for the course ASD 2456 taught by Professor Jojo during the Spring '10 term at Brooklyn Law School.

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Chapter_2_09 - Chapter 2 Problems 2.3 During a hard sneeze...

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