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University Physics: Mechanics Ch4. TWO- AND THREE-DIMENSIONAL MOTION Lecture 4 Dr.-Ing. Erwin Sitompul 2013

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4/2 Erwin SitompulUniversity Physics: Mechanics Solution of Homework 3: The Beetles Starting point 1 st run, 0.5 m 2 nd run, 0.8 m 1 st run, 1.6 m 2 nd run, ? New location

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4/3 Erwin SitompulUniversity Physics: Mechanics Thus, the second run of the green beetle corresponds to the vector Solution of Homework 3: The Beetles New location Starting point A B C D → → → →

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4/4 Erwin SitompulUniversity Physics: Mechanics Solution of Homework 3: The Beetles (a) The magnitude of the second run? (b) The direction of the second run? N E S W D → The direction of the second run is 79.09° south of due east or 10.91° east of due south.

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4/5 Erwin SitompulUniversity Physics: Mechanics Moving in Two and Three Dimensions In this chapter we extends the material of the preceding chapters to two and three dimensions. Position, velocity, and acceleration are again used, but they are now a little more complex because of the extra dimensions.

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4/6 Erwin SitompulUniversity Physics: Mechanics Position and Displacement One general way of locating a particle is with a position vector r, The coefficients x, z, and y give the particle’s location along the coordinate axes and relative to the origin. The following figure shows a particle with position vector In rectangular coordinates, the position is given by (–3 m, 2 m, 5 m). →

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4/7 Erwin SitompulUniversity Physics: Mechanics Position and Displacement As a particle moves, its position vector changes in a way that the vector always extends from the origin to the particle. If the position vector changes from r 1 to r 2, then the particle’s displacement delta is: →→

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4/8 Erwin SitompulUniversity Physics: Mechanics Average Velocity and Instantaneous Velocity If a particle moves through a displacement Δr in a time interval Δt, then its average velocity v avg is: The equation above can be rewritten in vector components as: → →

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4/9 Erwin SitompulUniversity Physics: Mechanics The particle’s instantaneous velocity v is the velocity of the particle at some instant. The direction of instantaneous velocity of a particle is always tangent to the particle’s path at the particle’s position. Average Velocity and Instantaneous Velocity →

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4/10 Erwin SitompulUniversity Physics: Mechanics Average Velocity and Instantaneous Velocity Writing the last equation in unit-vector form: This equation can be simplified by rewriting it as: where the scalar components of v are: The next figure shows a velocity vector v and its scalar x and y components. Note that v is tangent to the particle’s path at the particle’s position. → → →

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4/11 Erwin SitompulUniversity Physics: Mechanics The figure below shows a circular path taken by a particle. If the instantaneous velocity of the particle at a certain time is v = 2i – 2j m/s, through which quadrant is the particle currently moving when it is traveling (a) clockwise (b)counterclockwise (a) clockwise(b) counterclockwise First quadrant Third quadrant Average Velocity and Instantaneous Velocity ^ ^ →

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4/12 Erwin SitompulUniversity Physics: Mechanics When a particle’s velocity changes from v 1 to v 2 in a time interval Δt, its average acceleration a avg during Δt is: If we shrink Δt to zero, then a avg approaches the instantaneous acceleration a ; that is: Average and Instantaneous Acceleration →→ → → →

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4/13 Erwin SitompulUniversity Physics: Mechanics We can rewrite the last equation as where the scalar components of a are: Acceleration of a particle does not have to point along the path of the particle Average and Instantaneous Acceleration →

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4/14 Erwin SitompulUniversity Physics: Mechanics A particle with velocity v 0 = –2i + 4j m/s at t = 0 undergoes a constant acceleration a of magnitude a = 3 m/s 2 at an angle 130° from the positive direction of the x axis. What is the particle’s velocity v at t = 5 s? Solution: Thus, the particle’s velocity at t = 5 s is Average and Instantaneous Acceleration ^ → → ^ → At t = 5 s,

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4/15 Erwin SitompulUniversity Physics: Mechanics Homework 4: The Plane A plane flies 483 km west from city A to city B in 45 min and then 966 km south from city B to city C in 1.5 h. From the total trip of the plane, determine: (a) the magnitude of its displacement; (b) the direction of its displacement; (c) the magnitude of its average velocity; (d) the direction of its average velocity; (e) its average speed.

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4/16 Erwin SitompulUniversity Physics: Mechanics Homework 4A: The Turtle A turtle starts moving from its original position with the speed 10 cm/s in the direction 25° north of due east for 1 minute. Afterwards, it continues to move south for 2 m in 8 s. From the total movement of the turtle, determine: (a) the magnitude of its displacement; (b) the direction of its displacement; (c) the magnitude of its average velocity; (d) the direction of its average velocity; (e) its average speed.

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4/17 Erwin SitompulUniversity Physics: Mechanics Homework 4B: Shopping Trip 1.A shopper at a supermarket follows the path indicated by vectors S, H, O, and P in the figure. Given that the vectors have magnitudes S = 51 ft, H = 45 ft, O = 35 ft, and P = 13 ft. For all his movement, the shopper requires 8.5 min. Find: → →→ → (a)the magnitude of his displacement; (b)the direction of his displacement; (c)the magnitude of his average velocity; (d)the direction of his average velocity; (e)his average speed. 2.Compute A + B + C and express the result in magnitude-angle notation (polar coordinate). → →→

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