(Just need the practice problem worked out) the other picture have the info needed for the problems. Answer is provided in the text. 38CHAPTER 2 Motion Along a Straight LinePart (b): The change in velocity divided by the time intervalgives the average acceleration in each interval. The time interval isAr = 3.00 s-1.00 s= 2.00 s, soU2r- Ulx 4.00 m/s12-112.00 sPart (c): We use the same procedure to approximate the instantaneousacceleration at x₁ by calculating the average acceleration over a very shorttime period, 0.100 s. When At = 0.100 s, t₂ = 1.10 s, and it follows thatV2r = 60.0 m/s + (0.500 m/s³) (1.10 s)² = 60.605 m/s,Au, = 0.105 m/s,ax=dav..xAuxAt==0.105 m/s0.100 s=1.05 m/s².= 2.00 m/s².REFLECT If we repeat the calculation in part (c) for At = 0.01 s andAt = 0.001 s, we get a, 1.005 m/s² and a, 1.0005 m/s², respec-tively (although we aren't entitled to this many significant figures). AsAt gets smaller and smaller, the average acceleration gets closer andcloser to 1.00 m/s². We conclude that the instantaneous acceleration att = 1.00 s is 1.00 m/s².Practice Problem: What is the car's average acceleration betweent₁ = 2.00 s and t2 = 5.00 s? Answer: 3.5 m/s².As with average and instantaneous velocity, we can gain added insight into the conceptsof average and instantaneous acceleration by plotting a graph with velocity U, on the verticalaxis and time t on the horizontal axis. Figure 2.15 shows such a graph for the race car fromFigure 2.13. Notice that now we are plotting velocity versus time, not position versus time asbefore. If we draw a line between any two points on the graph, such as A and B (correspondingto a displacement from x₁ to x₂ and to the time interval At = 1₂equals the average acceleration over that interval. If we then take point B and move it closerti), the slope of that lineand closer to point A, the slope of the line AB approaches the slope of the line that is tanthe curve at point A. Accordingly- .no necessarily true that the object is speeding up.If a < 0, then it is not necessarily true that the object is slowing down.If ax = 0, then it is not necessarily true that the velocity of the object is also zero.The instantaneous acceleration at any point on a graph of velocity as a function oftime is the slope of the tangent to the curve at that point.EXAMPLE 2.4 Average and instantaneous accelerationsNow let's examine the differences between average and instantaneous accelerations. Suppose that, at anytime t, the velocity v of the car in Figure 2.13 is given by the equation vx = 60.0 m/s + (0.500 m/s³)². Inthis equation, v, has units of m/s and t has units of s. Note that the numbers 60.0 and 0.500 must have theunits shown in order for this equation to be dimensionally consistent. (a) Find the change in velocity of thecar in the time interval between t₁ = 1.00 s and t2 = 3.00 s. (b) Find the average acceleration in this timeinterval. (c) Estimate the instantaneous acceleration at time t₁ = 1.00 s by taking At = 0.10 s.SOLUTIONSET UP Figure 2.14 shows the diagram we use to establish a coordi-nate system and organize our known and unknown information.a₁₁x = ?X₁t₁ = 1.00 sKAvx = ?=?Gav, xA FIGURE 2.14 Our sketch for this problem.t₂ = 3.00 s✈→XO'XOVideo Tutor SolutionSOLVE Part (a): We first find the velocity at each time by substitutingeach value of t into the equation. From these values, we can find thechange in velocity in the interval. Thus, at time t₁ = 1.00 s,Vlx = 60.0 m/s + (0.500 m/s) (1.00 s)² = 60.5 m/s.At time t₂ = 3.00 s,V2x = 60.0 m/s + (0.500 m/s³) (3.00 s)² = 64.5 m/s.The change in velocity Au, is thenAUx = U2xVlx = 64.5 m/s60.5 m/s = 4.00 m/s.CONT

Name: (Just need the practice problem worked out) the other picture have the
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Description: (Just need the practice problem worked out) the other picture have the info needed for the problems. Answer is provided in the text. 38CHAPTER 2 Motion Along a Straight LinePart (b): The change in velocity divided by the time intervalgives the averag