Question

From t = 0 to t = 4.11 min, a man stands still, and from t = 4.11 min to t = 8.22 min, he walks briskly in a straight line at a constant speed of 2.98 m/s. What are (a) his average velocity vavg and (b) his average acceleration aavg in the time interval 1.00 min to 5.11 min? What are (c) vavg and (d) aavg in the time interval 2.00 min to 6.11 min?

Answer 1

Here ,

a) average velocity = total displacement/total time taken

average velocity = (4.11 * 60 * 2.98)/(8.22 * 60)

average velocity = 1.49 m/s

b)

average acceleration = change in velocity/time taken

average acceleration = (2.98 - 0)/(60 * (5.11 -1 ))

average acceleration = 0.0121 m/s^2

c)

for the interval between 2 min to 6.11

average velocity = total displacement/total time taken

average velocity = ((6.11 - 4.11) * 60 * 2.98)/(4.11 * 60)

average velocity = 1.45 m/s

d)

average acceleration = change in velocity/time taken

average acceleration = (2.98 - 0)/(60 * (6.11 - 2))

average acceleration = 0.0121 m/s^2