Pre-Calc Homework Solutions 291

Pre-Calc Homework Solutions 291 - Section 7.1 7. (a) Right...

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Unformatted text preview: Section 7.1 7. (a) Right when v(t) i.e., when 0 cos t cos t t 0, which is when cos t 3 or 2 2 2 291 0, 11. (a) v(t) C1 a(t) dt v(0) 90. 32(3) 32 dt 32t C1, where t 2 . Left when 0, i.e., when 0, i.e., when t 2 3 . Stopped when 2 3 or . 2 2 t Then v(3) (b) s(t) C2 16t v(t) dt s(0) 2 90 90t 0: 45) 6 ft/sec. C2, where (b) Displacement 0 e 0 sin t cos t dt e sin t 2 0 16t2 0. Solve s(t) 2t( 8t 45 8 [e 0 e ] 2 0 90t 0 or t 0 (c) Distance 3 /2 /2 0 esin t cos t dt 2 0 3 /2 /2 esin t cos t dt when t 5.625 sec. e sin t cos t dt 1) e 1 e e sin t cos t dt 1 1 e The projectile hits the ground at 5.625 sec. 2 e (e 2e 4.7 (c) Since starting height Displacement (d) Max. Height s 0. ending height, 8. (a) Right when v(t) 0, which is when 0 t 3. Left: never, since v(t) is never negative. Stopped when t 0. 3 (b) Displacement 0 t 1 t 2 dt 1 ln (1 2 t 2) 3 0 1 [ln (10) 2 3 ln (1)] t 1 t2 ln 10 2 ln 10 2 3/2 1.15 1.15 0, 16 5.625 2 2 5.625 2 5.625 90 2 126.5625, and 253.125 ft. 5 5 15 15 15 24 24 4 4 4 23 cm 33 cm 11. 5 5 16. 24 8. (c) Distance 0 dt 2t Distance 12. Displacement 2(Max. Height) c 9. (a) v(t) v(t) t a(t) dt t C, and since v(0) 9 2(27) v(t) dt 0 c 4 4 2t 3/2. Then v(9) 63 mph. 13. Total distance 0 v(t) dt a (b) First convert units: t t 3/2 t 2t 3/2 mph mi/sec. Then 3600 1800 9 3/2 t t Distance dt 1800 0 3600 2 5/2 t t 9 27 9 0 0.06525 mi 7200 4500 0 800 500 14. At t At t At t 15. At t a, s b, s c, s s(0) 0 b v(t) dt v(t) dt 0 c s(0) s(0) 0 v(t) dt a, where 344.52 ft. 4 dv is at a maximum (the graph is steepest dt upward). (t 0 10. (a) Displacement sin t [(sin 4 t cos t 2) sin t dt 4 0 16. At t 2] t 4, 2) sin t dt 1.44952 m c, where 2 cos t 2 cos 4) dv is at a maximum (the graph is steepest dt upward). 17. Distance Area under curve 4 1 2 4 cos 4 1 2 Distance 4 (b) Because the velocity is negative for 0 for 2 t 0 2, positive (a) Final position Initial position 2 4 6; ends at x 6. (b) 4 meters , and negative for 2 t (t 2 Distance (t 4 2) sin t dt (t 2) sin t dt sin 2 2 cos 4 2 sin 2 2) 18. (a) Positive and negative velocities cancel: the sum of signed areas is zero. Starts and ends at x 2. (b) Distance meters Sum of positive areas 4(1 1) 4 [(2 sin 2) ( ( sin 4 sin 4 2 2)] 1.91411 m. 2 2 cos 4 ...
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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