02_InstSolManual_PDF_Part7

02_InstSolManual_PDF_Part7 - 0.81 m. x 5 v x 5 v x 5 311 km...

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

View Full Document Right Arrow Icon
2.26. Set Up: The acceleration is the slope of the graph of versus t . Solve: (a) Reading from the graph, at to the right and at to the left. (b) versus t is a straight line with slope The acceleration is constant and equal to to the left. (c) The graph of versus t is given in Figure 2.26. Figure 2.26 2.27. Set Up: Assume constant acceleration. and Let Solve: (a) and (b) 2.28. Set Up: and Let and (a) and (b) or 2.29. Set Up: Assume constant acceleration. Take the direction to be downward, in the direction of the motion of the capsule. (stops), and Solve: (a) and The minus sign tells us that is upward. (b) so Reflect: Since the speed decreases, and must be in opposite directions. 2.30. Set Up: Let be in his direction of motion. Assume constant acceleration. (a) and (b) Solve: (a) and Yes, the time required is larger than 5.0 s. (b) v x 5 v 0 x 1 a x t 5 0 1 1 49.0 m / s 2 21 5.0 s 2 5 245 m / s t 5 v x 2 v 0 x a x 5 993 m / s 2 0 49.0 m / s 2 5 20.3 s v x 5 v 0 x 1 a x t t 5 5.0 s a x 5 5 g 5 49.0 m / s 2 . v 0 x 5 0, 993 m / s, v x 5 3 1 331 m / s 2 5 1 x a x v x t 5 v x 2 v 0 x a x 5 0 2 86.4 m / s 2 4.6 3 10 3 m / s 2 5 18.7 ms v x 5 v 0 x 1 a x t a x a x 5 v x 2 2 v 0 x 2 2 1 x 2 x 0 2 5 0 2 1 86.4 m / s 2 2 2 1 0.81 m 2 52 4.61 3 10 3 m / s 2 52 470 g . v x 2 5 v 0 x 2 1 2 a x 1 x 2 x 0 2 x 5
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 0.81 m. x 5 v x 5 v x 5 311 km / h 5 86.4 m / s, 1 x x 5 1 2 1 33 ft / s 2 21 2.0 s 2 2 5 66 ft x 5 x 1 v x t 1 1 2 a x t 2 5 1 2 1 10 m / s 2 21 2.0 s 2 2 5 20 m, a x 5 1 10 m / s 2 2 1 3.281 ft 1 m 2 5 33 ft / s 2 a x 5 v x 2 v x t 5 20.1 m / s 2 2.0 s 5 10 m / s 2 v x 5 v x 1 a x t 20.1 m / s. v x 5 45 mph 5 t 5 2.0 s, v x 5 0, x 5 0. 1 m 5 3.281 ft. 1 mph 5 0.4470 m / s x 5 x 1 v x t 1 1 2 a x t 2 5 1 88 ft / s 21 3.50 s 2 1 1 2 1 6.3 ft / s 2 21 3.50 s 2 2 5 347 ft a x 5 v x 2 v x t 5 110 ft / s 2 88 ft / s 3.50 s 5 6.3 ft / s 2 . v x 5 v x 1 a x t x 5 0. t 5 3.50 s. v x 5 110 ft / s, v x 5 88 ft / s, a x O t 2 1.3 m/s 2 a x 1.3 cm / s 2 , 2 8.0 cm / s 6.0 s 5 2 1.3 cm / s 2 . v x v x 5 1.3 cm / s, t 5 7.0 s, v x 5 2.7 cm / s, t 5 4.0 s, v x a x Motion along a Straight Line 2-7...
View Full Document

This note was uploaded on 03/06/2009 for the course PHYS 114 taught by Professor Shoberg during the Spring '07 term at Pittsburg State Uiversity.

Ask a homework question - tutors are online