# ch02-p021 - 21. In this solution, we make use of the...

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

t c b == = 2 3 230 320 10 (. ) ) . m/s s. 2 3 For t = 0, x = x 0 = 0 and for t = 1.0 s, x = 1.0 m > x 0 . Since we seek the maximum, we reject the first root ( t = 0) and accept the second ( t = 1s). (d) In the first 4 s the particle moves from the origin to x = 1.0 m, turns around, and goes back to x (( . ) ( . ( . ) ( . 43 0 4 0 2 0 4 0 8 0 2 s) m. 23 3 =−= The total path length it travels is 1.0 m + 1.0 m + 80 m = 82 m. (e) Its displacement is Δ x = x 2 x 1 , where x 1 = 0 and x 2 = –80 m. Thus, 80 m x Δ=− . The velocity is given by v = 2 ct – 3 bt 2 = (6.0 m/s 2 ) t – (6.0 m/s 3 ) t 2 . (f) Plugging in t = 1 s, we obtain 2 (1 s) (6.0 m/s )(1.0 s) (6.0 m/s )(1.0 s) 0. v =− = (g) Similarly, 2 (2 s) (6.0 m/s )(2.0 s) (6.0 m/s )(2.0 s) 12m/s . v = (h) 2 (3 s) (6.0 m/s )(3.0 s) (6.0 m/s )(3.0 s) 36 m/s . v = (i) 2 (4 s) (6.0 m/s )(4.0 s) (6.0 m/s )(4.0 s) 72 m/s . v = The acceleration is given by a = dv / dt = 2 c – 6 b = 6.0 m/s 2 – (12.0 m/s 3 ) t . (j) Plugging in t = 1 s, we obtain 2 6.0 m/s (12.0 m/s )(1.0 s) 6.0 m/s . a = (k) 2 (2 s) 6.0 m/s
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 09/03/2009 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.

Ask a homework question - tutors are online