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Ch. 4 Solutions - CHAPTER 4 PROBLEM SOLUTIONS Problem 4. j...

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Ch. 4 Solutions

Ch. 4 Solutions - CHAPTER 4 PROBLEM SOLUTIONS Problem 4. j...

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CHAPTER 4 PROBLEM SOLUTIONS Problem 4. An airliner is flying at a velocity of 260 î m/s , when a wind gust gives it an acceleration of 0 . 38 î + 0 . 72 ˆ j m / s 2 for a period of 24 s. (a) What is its velocity at the end of that time? (b) By what angle has it been deflected from its original course? Solution (a) Equation 4-3 gives v = 260 î m / s + (0.38 î + 0.72 ˆ j )( m / s 2 24 s ) = ( 269 î + 17 . 3 ˆ j ) m/s. (b) Since v 0 is along the x -axis, the angular deflection is just tan ! 1 ( v y =v x ) = tan ! 1 (17.3 = 269) = 3.67 ° . Problem 6. An airplane heads northeastward down a runway, accelerating from rest at the rate of 2.1 m/s 2 . Express the plane’s velocity and position at t = 30 s in unit vector notation, using a coordinate system with x -axis eastward and y -axis northward, and with origin at the start of the plane’s takeoff roll. Solution Since the acceleration is constant and the airplane starts from rest ( v 0 = 0) at the origin ( r 0 = 0), Equations 4-3 and 4-4 give v ( t ) = a t and r ( t ) = 1 2 a t 2 . Both vectors are in the NE direction, parallel to a = ( 2 . 1 m / s 2 î cos 45 ° + ˆ j sin 45 ° ) = ( 1 . 48 m / s 2 î + ˆ j ). Thus, v ( 30 s) = ( 1 . 48 m / s 2 30 s î + ˆ j ) = ( 44 . 5 m / s )( î + ˆ j ), and r ( 30 s ) = 1 2 ( 1 . 48 m/s)(30 s) 2 ( î + ˆ j ) = ( 668 m) ( î + ˆ j ). (Note: ( î + ˆ j ) = is a unit vector in the NE direction.) Problem 13. Figure 4-27 shows a cathode-ray tube, used to display electrical signals in oscilloscopes and other scientific instruments. Electrons are accelerated by the electron gun, then move down the center of the tube at 2.0 ! 10 9 cm/s. In the 4.2-cm- long deflecting region they undergo an acceleration directed perpendicular to the long axis of the tube. The acceleration “steers” them to a particular spot on the screen, where they produce a visible glow. (a) What acceleration is needed to deflect the electrons through 15 ° , as shown in the figure? (b) What is the shape of an electron’s path in the deflecting region? Solution (a) With x-y axes as drawn on Fig. 4-27, the electrons emerge from the deflecting region with velocity v = v 0 î + at ˆ j , after a time t = x 0 , where x = 4.2 cm and v 0 = 2 ! 10 9 cm/s. The angle of deflection (direction of v ) is 15 ° , so tan15 ° = v y x = at 0 = ax 0 2 . Thus, a = v 0 2 tan15 ° = x = 2.55 ! 10 17 cm/s 2 (when values are substituted). (b) Since the acceleration is assumed constant, the electron trajectory is parabolic in the deflecting region.
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