Question

From t = 0 to t = 5.00 min, a man stands still, and from t = 5.00 min to t = 10.0 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 2.00 min to 8.00 min? What are (c) vavg and (d) aavg in the time interval 3.00 min to 9.00 min? (e) Sketch x versus t and v versus t, and indicate how the answers to (a) through (d) can be obtained from the graphs.

Answer 1

a) Vavg

2.2m/s x 60s/min = 132m/min
t0 to t5 = 0m/min
t5 to t10 = 132m/min
total dist= 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
dist 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

Answer 2

wow superman lives

Answer 3

thanks