From t = 0 to t = 4.67 min, a man stands still, and from t = 4.67 min to t = 9.34 min, he walks briskly in a straight line at a constant speed of 1.97 m/s. What are (a) his average velocity vavg and (b) his average acceleration aavg in the time interval 1.00 min to 5.67 min? What are (c) vavg and (d) aavg in the time interval 2.00 min to 6.67 min?

Question

anonymous · Answer

a) ((1.97)*1*60 )/(5.67-1) = 25.31049251 m/m^2 / 60 = 0.4218415418 m/s
b) ((25.31049251)/4.67)/60 = 0.0903 m/s (wrong answer?) 
c) ((1.97)*2*60)/(6.67-2) = 50.62098501 m/m^2 / 60 = 0.8436830835m/s
d) (50.62098501/4.67 ) /60 = 0.1806 m/s (wrong answer ?)