9 In this expression v 0 ms and a is the acceleration due to gravity The

# 9 in this expression v 0 ms and a is the acceleration

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v0= 0 m/s, and ais the acceleration due to gravity. The velocity vwindowat the top of the window is not given, but it can be obtained from the time of 0.20 s that it takes the tile to pass the window. SOLUTIONSolving Equation 2.9 for yand using the fact that v0= 0 m/s gives 222window0window22vvvyaa(1) The tile travels an additional displacement ywindow=1.6 m in traversing the window in a time t= 0.20 s. These data can be used in 12windowwindow2yvtat(Equation 2.8) to find the velocity vwindowat the top of the window. Solving Equation 2.8 for vwindowgives 222windowwindow21.6 m9.80 m/s0.20 s27.0 m/s220.20 syatvtUsing this value for vwindowin Equation (1), we obtain 22window27.0 m/s2.5 m229.80 m/svya 80KINEMATICS IN ONE DIMENSION ______________________________________________________________________________ 65. SSMREASONINGThe slope of a straight-line segment in a position-versus-time graph is the average velocity. The algebraic sign of the average velocity, therefore, corresponds to the sign of the slope. SOLUTIONa. The slope, and hence the average velocity, is positivefor segments Aand C, negativefor segment B, and zerofor segment D. b. In the given position-versus-time graph, we find the slopes of the four straight-line segments to be 1.25 km0 km6.3 km/h0.20 h0 h0.50 km1.25 km3.8 km/h0.40 h0.20 h0.75 km0.50 km0.63 km/h0.80 h0.40 h0.75 km0.75 km0 km/h1.00 h0.80 hABCDvvvv   ______________________________________________________________________________ 66. REASONINGOn a position-versus-time graph, the velocity is the slope. Since the object’s velocity is constant and it moves in the +xdirection, the graph will be a straight line with a positive slope, beginning at x= −16 m when t= 0 s. At t= 18 s, its position should be x= 16 m + 48 m = +32 m. Once the graph is constructed, the object’s velocity is found by calculating the slope of the graph: xvt. SOLUTIONThe position-versus-time graph for the motion is as follows: Time t(s) Position x(m) 0 3.0 6.0 9.0 12 15 18 40 32 24 16 8.0 0.0 −8.016 −24 Chapter 2 Problems 81The object’s displacement is +48 m, and the elapsed time is 18 s, so its velocity is 48 m 2.7 m/s18 s xvt 67. REASONING AND SOLUTIONThe average acceleration for each segment is the slope of that segment. 2A2B2C40 m/s0 m/s1.9 m/s21 s0 s40 m/s0 m/s0 m/s48 s21 s80 m/s0 m/s3.3 m/s60 s48 saaa  ______________________________________________________________________________ 68. REASONINGThe average velocity for each segment is the slope of the line for that segment. SOLUTIONTaking the direction of motion as positive, we have from the graph for segments A, B, and C, 1110.0 km – 40.0 km–2.010 km/h1.5 h – 0.0 h20.0 km – 10.0 km1.010 km/h2.5 h – 1.5 h40.0 km – 20.0 km40 km/h3.0 h – 2.5 hABCvvv______________________________________________________________________________ 69. REASONINGThe slope of the position-time graph is the velocity of the bus. Each of the three segments of the graph is a straight line, so the bus has a different constant velocity for each part of the trip: vA, vB, and vC. The slope of each segment may be calculated from Equation 2.2  #### You've reached the end of your free preview.

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