(a) The entire interval considered is Δt= 8 – 2 = 6 min which is equivalent to 360 s, whereas the sub-interval in which he is movingis only 853min180 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, vavgmsm/s=−=3960360110..(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, aavg2s=−=2203600 00611.(c) Now, the entire interval considered is Δt= 9 – 3 = 6 min (360 s again), whereas the sub-interval in which he is moving is Δt'min=−==9 54240 s ). His position at 3 mint=isx= 0 and his position at t= 9 min isxvt=Δ'(.)( )2 2 240528 m . Therefore, vavgmsm/s.=−=5280360147.(d) The man is at rest at t= 3 min and has velocity v= +2.2 m/s at t= 9 min. Consequently, aavg= 2.2/360 = 0.00611 m/s2just as in part (b). (e) The horizontal line near the bottom of this x-vs-tgraph 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.