Business Calc Homework w answers_Part_59

Business Calc Homework w answers_Part_59 - 291 Section 7.1...

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7. (a) Right when v ( t ) . 0, which is when cos t . 0, i.e., when 0 # t , } p 2 } or } 3 2 p } , t # 2 p . Left when cos t , 0, i.e., when } p 2 } , t , } 3 2 p } . Stopped when cos t 5 0, i.e., when t 5 } p 2 } or } 3 2 p } . (b) Displacement 5 E 2 p 0 e sin t cos t dt 5 3 e sin t 4 5 [ e 0 2 e 0 ] 5 0 (c) Distance 5 E 2 p 0 ) e sin t cos t ) dt 5 E p /2 0 e sin t cos t dt 1 E 3 p /2 p /2 2 e sin t cos t dt 1 E 2 p 3 p /2 e sin t cos t dt 5 ( e 2 1) 1 1 e 2 } 1 e } 2 1 1 1 2 } 1 e } 2 5 2 e 2 } 2 e } < 4.7 8. (a) Right when v ( t ) . 0, which is when 0 , t # 3. Left: never, since v ( t ) is never negative. Stopped when t 5 0. (b) Displacement 5 E 3 0 } 1 1 t t 2 } dt 5 3 } 1 2 } ln (1 1 t 2 ) 4 5 } 1 2 } [ln (10) 2 ln (1)] 5 } ln 2 10 } < 1.15 (c) Distance 5 E 3 0 } 1 1 t t 2 } dt 5 } ln 2 10 } < 1.15 9. (a) v ( t ) 5 E a ( t ) dt 5 t 1 2 t 3/2 1 C , and since v (0) 5 0, v ( t ) 5 t 1 2 t 3/2 . Then v (9) 5 9 1 2(27) 5 63 mph. (b) First convert units: t 1 2 t 3/2 mph 5 } 36 t 00 } 1 } 1 t 8 3 0 /2 0 } mi/sec. Then Distance 5 E 9 0 1 } 36 t 00 } 1 } 1 t 8 3 0 /2 0 } 2 dt 5 3 } 72 t 0 2 0 } 1 } 4 t 5 5 0 /2 0 } 4 5 31 } 8 9 00 } 1 } 5 2 0 7 0 } 2 2 0 4 5 0.06525 mi 5 344.52 ft. 10. (a) Displacement 5 E 4 0 ( t 2 2) sin tdt 5 3 sin t 2 t cos t 1 2 cos t 4 5 [(sin 4 2 4 cos 4 1 2 cos 4) 2 2] < 2 1.44952 m (b) Because the velocity is negative for 0 , t , 2, positive for 2 , t , p , and negative for p , t # 4, Distance 5 E 2 0 2 ( t 2 2) sin t dt 1 E p 2 ( t 2 2) sin t dt 1 E 4 p 2 ( t 2 2) sin t dt 5 [(2 2 sin 2) 1 ( p 2 sin 2 2 2) 1 ( p 1 2 cos 4 2 sin 4 2 2)] 5 2 p 1 2 cos 4 2 2 sin 2 2 sin 4 2 2 < 1.91411 m. 11. (a) v ( t ) 5 E a ( t ) dt 5 E 2 32 dt 52 32 t 1 C 1 , where C 1 5 v (0) 5 90. Then v (3) 32(3) 1 90 6 ft/sec. (b) s ( t ) 5 E v ( t ) dt 16 t 2 1 90 t 1 C 2 , where C 2 5 s (0) 5 0. Solve s ( t ) 5 0: 2 16 t 2 1 90 t 5 2 t ( 2 8 t 1 45) 5 0 when t 5 0 or t 5 } 4 8 5 } 5 5.625 sec. The projectile hits the ground at 5.625 sec. (c) Since starting height 5 ending height, Displacement 5 0. (d) Max. Height 5 s 1 } 5.6 2 25 } 2 16 1 } 5.6 2 25 } 2 2 1 90 1 } 5.6 2 25 } 2 5 126.5625, and Distance 5 2(Max. Height) 5 253.125 ft. 12. Displacement 5 E c 0 v ( t ) dt 4 1 5 2 24 23 cm 13. Total distance 5 E c 0 ) v ( t ) ) dt 5 4 1 5 1 24 5 33 cm 14. At t 5 a , s 5 s (0) 1 E a 0 v ( t ) dt 5 15 2 4 5 11. At t 5 b , s 5 s (0) 1 E b 0 v ( t ) dt 5 15 2 4 1 5 5 16. At t 5 c , s 5 s (0) 1 E c 0 v ( t ) dt 5 15 2 4 1 5 2 24 8. 15. At t 5 a , where } d d v t } is at a maximum (the graph is steepest upward). 16. At t 5 c , where } d d v t } is at a maximum (the graph is steepest upward). 17. Distance 5 Area under curve 5 4 1 } 1 2 } ? 1 ? 2 2 5 4 (a) Final position 5 Initial position 1 Distance 5 2 1 4 5 6; ends at x 5 6. (b) 4 meters 18. (a) Positive and negative velocities cancel: the sum of signed areas is zero. Starts and ends at x 5 2. (b) Distance 5 Sum of positive areas 5 4(1 ? 1) 5 4 meters 4 0 9 0 3 0 2 p 0 Section 7.1 291
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19. (a) Final position 5 2 1 E 7 0 v ( t ) dt 5 2 2 } 1 2 } (1)(2) 1 } 1 2 } (1)(2) 1 1(2) 1 } 1 2 } (2)(2) 2 } 1 2 } (2)(1) 5 5; ends at x 5 5.
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Business Calc Homework w answers_Part_59 - 291 Section 7.1...

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