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CHAPTER 4 PROBLEM SOLUTIONS Problem 4. An airliner is flying at a velocity of 260 m/s , when a wind gust gives it an acceleration of . 38 + . 72 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 m / s + (0.38 + 0.72 j )( m / s 2 )( 24 s ) = ( 269 + 17 . 3 j ) m/s. (b) Since v 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 planes 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 planes takeoff roll. Solution Since the acceleration is constant and the airplane starts from rest ( v = 0) at the origin ( r = 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 )( cos 45 + j sin 45 ) = ( 1 . 48 m / s 2 )( + j ). Thus, v ( 30 s) = ( 1 . 48 m / s 2 )( 30 s )( + j ) = ( 44 . 5 m / s )( + j ), and r ( 30 s ) = 1 2 ( 1 . 48 m/s)(30 s) 2 ( + j ) = ( 668 m) ( + j ). (Note: ( + j ) = 2 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 electrons 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 + at j , after a time t = x =v , where x = 4.2 cm and v = 2 ! 10 9 cm/s. The angle of deflection (direction of v ) is 15 , so tan15 = v y =v x = at =v = ax =v 2 . Thus, a = v 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. CHAPTER 4 71 Problem 14. You toss an apple horizontally at 8.7 m/s from a height of 2.6 m. Simultaneously, you drop a peach from the same height. How long does each take to reach the ground? ... View Full Document

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