chap3 - CHAPTER 3 THE VECTOR DESCRIPTION OF MOTION...

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Unformatted text preview: CHAPTER 3 THE VECTOR DESCRIPTION OF MOTION ActivPhysics can help with these problems: Activity 4.1 Section 3-2: Vector Arithmetic Problem 1. You walk west 220 m, then north 150 m. What are the magnitude and direction of your displacement vector? Solution The triangle formed by the two displacement vectors and their sum is a right triangle, so the Pythagorean Theorem gives the magnitude C = A 2 + B 2 = radicalbig (220 m) 2 + (150 m) 2 = 266 m, and the basic defi nition of the tangent gives = tan 1 (150 m / 220 m) = 34 . 3 . The direction of C can be specified as 34 . 3 N of W, or 55.7 W of N, or by the azimuth 304.3 (CW from N), etc. Problem 1 Solution. Problem 2. An ion in a mass spectrometer (a device that sorts atomic-size particles) follows a semicircular path of radius 15.2 cm. What are (a) the distance it travels and (b) the magnitude of its displacement? Solution (a) The length of the semicircle is 1 2 (2 r ) = r = (15 . 2 cm) = 47 . 8 cm . (b) The magnitude of the displacement vector, from the start of the semicircle to its end, is just a diameter, or 2(15 . 2 cm) = 30 . 4 cm . Problem 3. A migrating whale follows the west coast of Mexico and North America toward its summer home in Alaska. It first travels 360 km due northwest to just off the coast of Northern California and then turns due north and travels 400 km toward its destination. Determine graphically the magnitude and direction of its displacement vector. Solution We can find the magnitude and direction of the vector sum of the two displacements either using geometry and a diagram, or by adding vector components. From the law of cosines: C = radicalbig A 2 + B 2- 2 AB cos = radicalbig (360 km) 2 + (400 km) 2- 2(360 km)(400 km) cos 135 = 702 km . From the law of sines: C/ sin = B/ sin , or =sin 1 parenleftbigg B sin C parenrightbigg = sin 1 bracketleftbiggparenleftbigg 400 m 702 m parenrightbigg sin 135 bracketrightbigg = 23 . 7 . The direction of C can be specified as 45 + 23 . 7 = 68 . 7 N of W, or 180 68 . 7 = 111 CCW from the x-axis (east) in the illustration. Problem 3 Solution. In a coordinate system with x-axis east and y-axis north, the first displacement is 360 km ( cos135 + sin 135 ) and the second simply 400 km . Their sum is ( 255 + 255 + 400 ) km = ( 255 + 655 ) km, 2 CHAPTER 3 which is the total displacement. Its magnitude is radicalbig ( 255) 2 + (655) 2 km / 702 km and its direction (measured CCW from the x-axis) is x = cos 1 ( 255 km / 702 km) = 111 , as above. (Note that since C y > 0 and C x < , x is in the second quadrant.) Problem 4. A citys streets are laid out with its north-south blocks twice as long as its east-west blocks. You walk 8 blocks east and 3 blocks north. Determine (a) the total distance youve walked and (b) the magnitude of your displacement vector. Express inmagnitude of your displacement vector....
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This note was uploaded on 07/10/2008 for the course PHYS 2A taught by Professor Hicks during the Summer '07 term at UCSD.

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chap3 - CHAPTER 3 THE VECTOR DESCRIPTION OF MOTION...

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