From t = 0 to t = 3.68 min, a man stands still, and from t = 3.68 min to t = 7.3

Question

From t = 0 to t = 3.68 min, a man stands still, and from t = 3.68 min to t = 7.36 min, he walks briskly in a straight line at a constant speed of 2.20 m/s. What are (a) his average velocity vavg and (b) his average acceleration aavg in the time interval 1.00 min to 4.68 min? What are (c) vavg and (d) aavg in the time interval 2.00 min to 5.68 min? Note: Please put your answers in a) b) c) notation so I can understand what you're typing. If possible, please show your work -- I'd like to understand how you got the answer instead of just the answers.

Explanation / Answer

a) Vavg=? 2.2m/s x 60s/min = 132m/min t0 to t5 = 0m/min t5 to t10 = 132m/min total distance = 132m/min x 5min = 660m Vavg=(total distance /total time) = (660m/10min) = 66m/min b) Aavg t=2 to t=8 =? Aavg=(vt8 - vt2)/t = (132m/min-0)/6min = 22m/min^2 c) Vavg t3-t9 = ? distance t3-t9 = 132m/min x 4min = 528m Vavg t3-t9 = (528m/6min) = 88m/min d) Aavg t3-t9 = ? Vt3=0, Vt9=132m/min Aavg t3-t9 = (132m/min-0)/6min = 22m/min^2 hope this help ya..

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