What is the car’s average acceleration between t1= 2.00s and t2= 5.00s? Answer 3.5m/s^2Both images have background info to help solve the practice problem. w ittheera-arentInstantaneous acceleration plays an essential role in the laws of mechanics. From nowon, when we use the term acceleration, we will always mean instantaneous accelera-tion, not average acceleration.Units: m/s2Notes:. If a, > 0, then it is not necessarily true that the object is speeding up.. If a, < 0, then it is not necessarily true that the object is slowing down.. If a, = 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 r, the velocity v of the car in Figure 2.13 is given by the equation U = 60.0 m/s + (0.500 m/s³)². Inthis equation, u, 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.91, x = ?X1t₁ =1.00 sAvx = ?aav, x = ?A FIGURE 2.14 Our sketch for this problem.KX₂t₂ = 3.00 sVideo Tutor SolutionSOLVE Part (a): We first find the velocity 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,U1x = 60.0 m/s + (0.500 m/s³) (1.00 s)² = 60.5 m/s.At time t₂ = 3.00 s,U2x = 60.0 m/s + (0.500 m/s³) (3.00 s)² = 64.5 m/s.The change in velocity Au, is thenAUx = U2x U1x = 64.5 m/s 60.5 m/s = 4.00 m/s.CONTINUED CHAPTER 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 is1.00 s 2.00 s, soAr-3.00 s4.00 m/s2.00 sPart (c): We use the same procedure to approximate the instantaneousacceleration at .x, by calculating the average acceleration over a very shorttime period, 0.100 s. When Ar= 0.100 s, ₂= 1.10 s, and it follows thatU2= 60.0 m/s + (0.500 m/s³) (1.10 s)² = 60.605 m/s,Av,= 0.105 m/s,aAv.xAUXΔΙVzx - VLx0.105 m/s0.100 s= 1.05 m/s².2.15 A graph of velocity versus= motion of the race car in= 2.00 m/s².As with average and instantaneous velocity, we can gain added insight into the conceof average and instantaneous acceleration by plotting a graph with velocity U, on the vertaxis and time t on the horizontal axis. Figure 2.15 shows such a graph for the race car fiFigure 2.13. Notice that now we are plotting velocity versus time, not position versus timbefore. If we draw a line between any two points on the graph, such as A and B (correspon-to a displacement from x₁ to x₂ and to the time interval At = t₂t₁), the slope of thatequals the average acceleration over that interval. If we then take point B and move it cand closer to point A, the slope of the line AB approaches the slope of the line that is tang-the curve at point A. Accordingly, we can interpret a graph of velocity versus time as folThe average acceleration between any two points on a graph of velocityversus time equals the slope of a line connecting those points.U2xThe instantaneous acceleration at any point on the graph equals the slope cthe line tangent to the curve at that point.Ux000 anodrAUlx ----0REFLECT If we repeat the calculation in part (c) for Ar= 0.01 s andAr = 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). AsAr gets smaller and smaller, the average acceleration gets closer andcloser to 1.00 m/s². We conclude that the instantaneous acceleration at= 1.00 s is 1.00 m/s².Practice Problem: What is the car's average acceleration between₁ = 2.00 s and 2 = 5.00 s? Answer: 3.5 m/s².EPTUAL ANALYSIS 2.3in a two-story buildingelevator in a building that has only two floors. As the eleva-stop so that you can get off, you are accelerating upwardinformation whicht1Slope average acceleration12-1₁ = AtU2x U₁x = AUxSlope of tangent= instantaneous acceleration at As1₂tSOLUTION Because the elevator is slowing down, we kn-elevator's velocity vector points in the opposite direction as ition vector. Since the elevator's 000

Name: What is the car’s average acceleration between t1= 2.00s and t2= 5.00s
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Description: What is the car’s average acceleration between t1= 2.00s and t2= 5.00s? Answer 3.5m/s^2Both images have background info to help solve the practice problem. w ittheera-arentInstantaneous acceleration plays an essential role in the laws of mechanics. F