Only Practice problem 3.1Second photo has background info for the problem. val At(3.3)tor AFimit isy areneousOcity8 inSOLUTIONSET UP Figure 3.3 shows our sketch for this problem. We see thatAx = 7.0 m4.0 m = 3.0 m, Ay = 6.0 m - 2.0 m = 4.0 m, andAt = 0.50 s.SOLVE Part (a): To find the components of av, we use their defini-tions (Equations 3.2):Vav, x=Vav, y=AxAtAyAt|av| ==AyAx-18 = tan3.0 m0.50 sPart (b): The magnitude of av is obtained from the Pythagorean theorem:|av| = √Uav. x² + Uav, y4.0 m√(6.0 m/s)² + (8.0 m/s)² = 10.0 m/s.0.50 sThe direction of av is most easily described by its angle, measuredcounterclockwise from the positive x axis. Calling this angle 0, we have5.0 m0.50 s= 6.0 m/s,= tan6 = tan-1 av, yVav, xAlternative Solution: Alternatively, the magnitude of ay is the dis-tance between the points (4.0 m, 2.0 m) and (7.0 m, 6.0 m)-thatis, 5.0 m (found using Pythagoras's theorem)-divided by the timeinterval (0.5 s):= 8.0 m/s.= tan8.0 m/s6.0 m/sSince the direction of day is the same as the direction of the displace-ment AF between the two points, we can calculate it from Ax and Ayrather the from Uav, x and Uav, y:= 10.0 m/s.4.0 m3.0 m= 53°.gos sudents= 53°.ay points from P, toward P₂. It doesn't matterto moire found mathematically.how long you make it; its magnitude will bebumsidoy (m)6.03.2 Acceleration in a PlaneΔΥAy 4.0m Vavy At2.0OP₁0†₁-2.0 si4.022.5sP₂A FIGURE 3.3 Our sketch for this problem.gado ai nobonto mua vota s7.0Ax-3.0mAxAt→x (m)67REFLECT Be sure you understand the relationship between the twosolutions to part (b). In the first, we calculated the magnitude and direc-tion of Jay from the components of this vector quantity. In the alterna-tive solution, we used the fact that the average velocity ay and thedisplacement AF have the same direction.Practice Problem: Suppose you reverse the car's motion, so thatit retraces its path in the opposite direction in the same time. Findthe components of the average velocity of the car and the magnitudeand direction of the average velocity. Answers: -6.0 m/s, -8.0 m/s,-10.0 m/s, 233°. 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°-

Name: Only Practice problem 3.1Second photo has background info for the prob
Uploaded: 2024-03-05T11:56:34.000Z
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Description: Only Practice problem 3.1Second photo has background info for the problem. val At(3.3)tor AFimit isy areneousOcity8 inSOLUTIONSET UP Figure 3.3 shows our sketch for this problem. We see thatAx = 7.0 m4.0 m = 3.0 m, Ay = 6.0 m - 2.0 m = 4.0 m, andAt =