Business Calc Homework w answers_Part_59

625 2 2 distance 115 ending height 5625 2 5625 90 2

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Unformatted text preview: ) v(t) a(t) dt v(t) t 2t 3/2 C, and since v(0) 2t 3/2. Then v(9) t 9 2(27) 0, 63 mph. 12. Displacement v(t) dt 4 5 24 23 cm 0 c 13. Total distance v(t) dt 4 5 24 33 cm v(t) dt 15 4 11. v(t) dt 15 4 5 16. v(t) dt 15 4 5 24 0 a (b) First convert units: 14. At t 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 0 800 500 344.52 ft. a, s s(0) 0 b At t b, s s(0) 0 c At t c, s s(0) 8. 0 15. At t a, where dv is at a maximum (the graph is steepest dt upward). 4 10. (a) Displacement (t 2) sin t dt 0 sin t 16. At t 4 t cos t 2 cos t 4 cos 4 2 cos 4) 2] 1.44952 m 17. Distance (b) Because the velocity is negative for 0 for 2 t , and negative for t t 2, positive 4, 2 Distance (t 0 2) sin t dt [(2 sin 2) ( ( 2 2 cos 4 (t 2) sin t dt 2 4 (t 2) sin t dt sin 2 2 cos 4 2 sin 2 dv is at a maximum (the graph is steepest dt upward). 0 [(sin 4 c, where sin 4 2)] 2 1.91411 m. 4 1 2 (a) Final position Initial position 2 4 6; ends at x 6. 12 4 Distance (b) 4 meters 18. (a) Positive and negative velocities cancel: the sum of signed areas is z...
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This document was uploaded on 10/31/2011 for the course MAC 2311 at University of Florida.

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