Only practice problem 3.2Second photo had background info for problem. 70CHAPTER 3 Motion in a PlaneSOLUTIONSET UP Figure 3.7 shows our sketch.SOLVE Figure 3.6a outlines the relationships we'll use.Part (a): To find the components of average acceleration, we needthe components of the change in velocity, Au, and Avy, and thetime interval, At = 0.5 s. The change in u, is Av, U2x - Ulx =(4.0 m/s 1.0 m/s) = 3.0 m/s, so the x component of average accel-eration in the interval At = 0.5 s isaav, x=AuxAt3.0 m/s0.5 s= 6.0 m/s².The change in u, is zero, so dav, y in this interval is also zero.Part (b): The vector day has only an x component. The vector points inthe +x direction and has magnitude 6.0 m/s².REFLECT We can always represent a vector quantity (such as dis-placement, velocity, or acceleration) either in terms of its componentsor in terms of its magnitude and direction.CONCEPTUAL ANALYSIS 3.1A visitor at apicnicAn ant crawls at a constant speed on top of a square blanket that you havelaid out for a picnic. You have picked a flat, horizontal lawn for your picnicand oriented the blanket so that one side of it runs porother side runs east-west.y (m)t₂ = 2.5 sP₂V2y = 3.0 m/sV2x = 4.0 m/sV₁y = 3.0 m/sP₁--t₁ = 2.0 S V₁x = 1.0 m/sA FIGURE 3.7 Our sketch for this problem.x (m)Practice Problem: Suppose that as the car continues to move, it has vel-ocity components Ux = 3.5 m/s and uy = -1.0 m/s at time 13 = 3.0 s.What are the magnitude and direction of the average acceleration bet-ween t2 = 2.0 s and t3 = 3.0 s? Answers: 4.7 m/s, -58° from +x.the ant's displacement points in the southwest direction, so the averagevelocity must point to the southwest as well Chapter 3: Practices ProblemsName:G#Page 66- Practice Problem 3.1Suppose you reverse the car's motion, so that it retraces itspath in the opposite direction in the same time. Find thecomponents of the average velocity of the car and themagnitude and direction of the average velocity. Answer:-6.0m/s, -8.0m/s, -10.0m/s, 233⁰0Page 69- Practice Problem 3.2Suppose that as the car continue to move, it has velocitycomponents v= 3.5 m/s and v, 1.0 m/s at time t,= 3.0s.What are the magnitude and direction of the averageacceleration between t₂=2.0 s and t₂= 3.0s? Answer: 4.7m/s,-58⁰ from +x.→ Page 73- Practice Problem 3.3If air resistance is ignored, what initial speed is requiredfor a range of 20m? Answer: 36.1 m/s.Page 74- Practice problem 3.4If the ball could continue to travel below its original level(through an appropriately shaped hole in the ground), thennegative values of y corresponding to times greater than6.04s would be possible. Compare the ball's position andvelocity 8.00s after the start of its flight. Answers: x=178m.y=-76.8m, v, 22.2m/s, v,=-48.8 m/s.Page 75- Practice Problem 3.5Show the range is the same when the launch angle is 30° aswhen it is 60° and that the range for any launch angle isthe same as for the complementary angle (90°-Answer: sinsin(180-), so sin2000 sin2(90°-

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Only practice problem 3.2Second photo had background info for problem. 70CHAPTER 3 Motion in a PlaneSOLUTIONSET UP Figure 3.7 shows our sketch.SOLVE Figure 3.6a outlines the relationships we'll use.Part (a): To find the components of average acceleration, we needthe components of the change in velocity, Au, and Avy, and thetime interval, At = 0.5 s. The change in u, is Av, U2x - Ulx =(4.0 m/s 1.0 m/s) = 3.0 m/s, so the x component of average accel-eration in the interval At = 0.5 s isaav, x=AuxAt3.0 m/s0.5 s= 6.0 m/s².The change in u, is zero, so dav, y in this interval is also zero.Part (b): The vector day has only an x component. The vector points inthe +x direction and has magnitude 6.0 m/s².REFLECT We can always represent a vector quantity (such as dis-placement, velocity, or acceleration) either in terms of its componentsor in terms of its magnitude and direction.CONCEPTUAL ANALYSIS 3.1A visitor at apicnicAn ant crawls at a constant speed on top of a square blanket that you havelaid out for a picnic. You have picked a flat, horizontal lawn for your picnicand oriented the blanket so that one side of it runs porother side runs east-west.y (m)t₂ = 2.5 sP₂V2y = 3.0 m/sV2x = 4.0 m/sV₁y = 3.0 m/sP₁-->t₁ = 2.0 S V₁x = 1.0 m/sA FIGURE 3.7 Our sketch for this problem.>x (m)Practice Problem: Suppose that as the car continues to move, it has vel-ocity components Ux = 3.5 m/s and uy = -1.0 m/s at time 13 = 3.0 s.What are the magnitude and direction of the average acceleration bet-ween t2 = 2.0 s and t3 = 3.0 s? Answers: 4.7 m/s, -58° from +x.the ant's displacement points in the southwest direction, so the averagevelocity must point to the southwest as well Chapter 3: Practices ProblemsName:G#Page 66- Practice Problem 3.1Suppose you reverse the car's motion, so that it retraces itspath in the opposite direction in the same time. Find thecomponents of the average velocity of the car and themagnitude and direction of the average velocity. Answer:-6.0m/s, -8.0m/s, -10.0m/s, 233⁰0Page 69- Practice Problem 3.2Suppose that as the car continue to move, it has velocitycomponents v= 3.5 m/s and v, 1.0 m/s at time t,= 3.0s.What are the magnitude and direction of the averageacceleration between t₂=2.0 s and t₂= 3.0s? Answer: 4.7m/s,-58⁰ from +x.→ Page 73- Practice Problem 3.3If air resistance is ignored, what initial speed is requiredfor a range of 20m? Answer: 36.1 m/s.Page 74- Practice problem 3.4If the ball could continue to travel below its original level(through an appropriately shaped hole in the ground), thennegative values of y corresponding to times greater than6.04s would be possible. Compare the ball's position andvelocity 8.00s after the start of its flight. Answers: x=178m.y=-76.8m, v, 22.2m/s, v,=-48.8 m/s.Page 75- Practice Problem 3.5Show the range is the same when the launch angle is 30° aswhen it is 60° and that the range for any launch angle isthe same as for the complementary angle (90°-Answer: sinsin(180-), so sin2000 sin2(90°-

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Question no.- 3.2
What are the magnitude and direction of average acceleration between t2=2.0 s and…

Page 69- Practice Problem 3.2 Suppose that as the car continue to move, it has velocity components v₂= 3.5 m/s and v, 1.0 m/s at time t,= 3.0s. What are the magnitude and direction of the average acceleration between t₂=2.0 s and t₂= 3.0s? Answer: 4.7m/s, -58⁰ from +x.