Pre-Calc Homework Solutions 291

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

## 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.

Ask a homework question - tutors are online