Question #cb25a

1 Answer
Feb 27, 2018

a) time interval 1, zero; time interval 2, 2.02 m/s; time interval 1+2, 1.01 m/s.
b) time interval 1, zero.

Explanation:

a) The average velocity #v_"avg"# in time interval 1 is zero because he is standing still. In time interval 2, it is 2.02 m/s because that was his constant speed.

To calculate the distance he walked, we first need the time in seconds.

#t = 4.96 cancel(min) * (60 s)/(1 cancel(min)) = 297.6 s#

The distance he walks in time interval 2 is

#s = v*t = 2.02 m/s * 297.6 s = 601.2 m #

To calculate #v_"avg"# from t = 0 to t = 9.92 min, I will use the total time from both time intervals
#t_"total" = 2*297.6 s = 595.2 s#

#v_"avg" = "total distance"/"total time" = (601.2 m)/(595.2 s) = 1.01 m/s#

The question is not clear about the time period for which it wants the #v_"avg"#. So I gave you #v_"avg"# for time interval 1, time interval 2, and time interval 1+2.

b) His average acceleration #a_"avg "#in the time interval 1 is zero because he is standing still.

I hope this helps,
Steve