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What is the car’s average acceleration between t1= 2.00s and t2= 5.00s? Answer 3.5m/s^2 Both images have background info to help solve the practice problem.

What is the car’s average acceleration between t1= 2.00s and t2= 5.00s? Answer 3.5m/s^2 Both images have background info to help solve the practice problem.

University Physics Volume 1
University Physics Volume 1
18th Edition
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:William Moebs, Samuel J. Ling, Jeff Sanny
Chapter6: Applications Of Newton's Laws
Section: Chapter Questions
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What is the car’s average acceleration between t1= 2.00s and t2= 5.00s? Answer 3.5m/s^2 Both images have background info to help solve the practice problem.
w it the era- are n t Instantaneous acceleration plays an essential role in the laws of mechanics. From now on, when we use the term acceleration, we will always mean instantaneous accelera- tion, not average acceleration. Units: m/s2 Notes: . 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 of time is the slope of the tangent to the curve at that point. EXAMPLE 2.4 Average and instantaneous accelerations Now let's examine the differences between average and instantaneous accelerations. Suppose that, at any time 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. In this equation, u, has units of m/s and t has units of s. Note that the numbers 60.0 and 0.500 must have the units shown in order for this equation to be dimensionally consistent. (a) Find the change in velocity of the car in the time interval between t₁ = 1.00 s and t₂ = 3.00 s. (b) Find the average acceleration in this time interval. (c) Estimate the instantaneous acceleration at time t₁ = 1.00 s by taking At = 0.10 s. SOLUTION SET UP Figure 2.14 shows the diagram we use to establish a coordi- nate system and organize our known and unknown information. 91, x = ? X1 t₁ =1.00 s Avx = ? aav, x = ? A FIGURE 2.14 Our sketch for this problem. K X₂ t₂ = 3.00 s Video Tutor Solution SOLVE Part (a): We first find the velocity each time by substituting each value of t into the equation. From these values, we can find the change 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 then AUx = U2x U1x = 64.5 m/s 60.5 m/s = 4.00 m/s. CONTINUED
Transcribed Image Text:w it the era- are n t Instantaneous acceleration plays an essential role in the laws of mechanics. From now on, when we use the term acceleration, we will always mean instantaneous accelera- tion, not average acceleration. Units: m/s2 Notes: . 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 of time is the slope of the tangent to the curve at that point. EXAMPLE 2.4 Average and instantaneous accelerations Now let's examine the differences between average and instantaneous accelerations. Suppose that, at any time 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. In this equation, u, has units of m/s and t has units of s. Note that the numbers 60.0 and 0.500 must have the units shown in order for this equation to be dimensionally consistent. (a) Find the change in velocity of the car in the time interval between t₁ = 1.00 s and t₂ = 3.00 s. (b) Find the average acceleration in this time interval. (c) Estimate the instantaneous acceleration at time t₁ = 1.00 s by taking At = 0.10 s. SOLUTION SET UP Figure 2.14 shows the diagram we use to establish a coordi- nate system and organize our known and unknown information. 91, x = ? X1 t₁ =1.00 s Avx = ? aav, x = ? A FIGURE 2.14 Our sketch for this problem. K X₂ t₂ = 3.00 s Video Tutor Solution SOLVE Part (a): We first find the velocity each time by substituting each value of t into the equation. From these values, we can find the change 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 then AUx = U2x U1x = 64.5 m/s 60.5 m/s = 4.00 m/s. CONTINUED
CHAPTER 2 Motion Along a Straight Line Part (b): The change in velocity divided by the time interval gives the average acceleration in each interval. The time interval is 1.00 s 2.00 s, so Ar-3.00 s 4.00 m/s 2.00 s Part (c): We use the same procedure to approximate the instantaneous acceleration at .x, by calculating the average acceleration over a very short time period, 0.100 s. When Ar= 0.100 s, ₂= 1.10 s, and it follows that U2= 60.0 m/s + (0.500 m/s³) (1.10 s)² = 60.605 m/s, Av,= 0.105 m/s, a Av.x AUX ΔΙ Vzx - VLx 0.105 m/s 0.100 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 conce of average and instantaneous acceleration by plotting a graph with velocity U, on the vert axis and time t on the horizontal axis. Figure 2.15 shows such a graph for the race car fi Figure 2.13. Notice that now we are plotting velocity versus time, not position versus tim before. 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 that equals the average acceleration over that interval. If we then take point B and move it c and 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 fol The average acceleration between any two points on a graph of velocity versus time equals the slope of a line connecting those points. U2x The instantaneous acceleration at any point on the graph equals the slope c the line tangent to the curve at that point. Ux 000 anodr A Ulx ---- 0 REFLECT If we repeat the calculation in part (c) for Ar= 0.01 s and Ar = 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). As Ar gets smaller and smaller, the average acceleration gets closer and closer 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 between ₁ = 2.00 s and 2 = 5.00 s? Answer: 3.5 m/s². EPTUAL ANALYSIS 2.3 in a two-story building elevator in a building that has only two floors. As the eleva- stop so that you can get off, you are accelerating upward information which t1 Slope average acceleration 12-1₁ = At U2x U₁x = AUx Slope of tangent = instantaneous acceleration at A s 1₂ t SOLUTION Because the elevator is slowing down, we kn- elevator's velocity vector points in the opposite direction as i tion vector. Since the elevator's 000
Transcribed Image Text:CHAPTER 2 Motion Along a Straight Line Part (b): The change in velocity divided by the time interval gives the average acceleration in each interval. The time interval is 1.00 s 2.00 s, so Ar-3.00 s 4.00 m/s 2.00 s Part (c): We use the same procedure to approximate the instantaneous acceleration at .x, by calculating the average acceleration over a very short time period, 0.100 s. When Ar= 0.100 s, ₂= 1.10 s, and it follows that U2= 60.0 m/s + (0.500 m/s³) (1.10 s)² = 60.605 m/s, Av,= 0.105 m/s, a Av.x AUX ΔΙ Vzx - VLx 0.105 m/s 0.100 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 conce of average and instantaneous acceleration by plotting a graph with velocity U, on the vert axis and time t on the horizontal axis. Figure 2.15 shows such a graph for the race car fi Figure 2.13. Notice that now we are plotting velocity versus time, not position versus tim before. 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 that equals the average acceleration over that interval. If we then take point B and move it c and 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 fol The average acceleration between any two points on a graph of velocity versus time equals the slope of a line connecting those points. U2x The instantaneous acceleration at any point on the graph equals the slope c the line tangent to the curve at that point. Ux 000 anodr A Ulx ---- 0 REFLECT If we repeat the calculation in part (c) for Ar= 0.01 s and Ar = 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). As Ar gets smaller and smaller, the average acceleration gets closer and closer 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 between ₁ = 2.00 s and 2 = 5.00 s? Answer: 3.5 m/s². EPTUAL ANALYSIS 2.3 in a two-story building elevator in a building that has only two floors. As the eleva- stop so that you can get off, you are accelerating upward information which t1 Slope average acceleration 12-1₁ = At U2x U₁x = AUx Slope of tangent = instantaneous acceleration at A s 1₂ t SOLUTION Because the elevator is slowing down, we kn- elevator's velocity vector points in the opposite direction as i tion vector. Since the elevator's 000
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