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

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Presentation on theme: "University Physics: Mechanics Ch4. TWO- AND THREE-DIMENSIONAL MOTION Lecture 4 Dr.-Ing. Erwin Sitompul 2013."— Presentation transcript:

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

2 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

3 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 → → → →

4 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.

5 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.

6 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). →

7 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: →→

8 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: → →

9 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 →

10 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. → → →

11 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 ^ ^ →

12 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 →→ → → →

13 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 →

14 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,

15 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.

16 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.

17 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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