# ch02-p022 - 22 We use Eq 2-2(average velocity and Eq...

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(a) The entire interval considered is Δ t = 8 – 2 = 6 min which is equivalent to 360 s, whereas the sub-interval in which he is moving is only 8 5 3min 180 s. t' Δ=−= = His position at t = 2 min is x = 0 and his position at t = 8 min is x vt == Δ ' (2.2)(180) 396 m = . Therefore, v avg m s m/s = = 396 0 360 110 .. (b) The man is at rest at t = 2 min and has velocity v = +2.2 m/s at t = 8 min. Thus, keeping the answer to 3 significant figures, a avg 2 s = = 22 0 360 0 00611 . (c) Now, the entire interval considered is Δ t = 9 – 3 = 6 min (360 s again), whereas the sub-interval in which he is moving is Δ t 'm i n =−= = 9 5 4 240 s ). His position at 3 min t = is x = 0 and his position at t = 9 min is xv t = Δ '( . ) ( ) 2 2 240 528 m . Therefore, v avg m s m/s. = = 528 0 360 147 . (d) The man is at rest at t = 3 min and has velocity v = +2.2 m/s at t = 9 min. Consequently, a avg = 2.2/360 = 0.00611 m/s 2 just as in part (b). (e) The horizontal line near the bottom of this x -vs- t graph represents the man standing at x = 0 for 0 t < 300 s and the linearly rising line for 300 t 600 s represents his
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## 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.

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