Question

Question #cb25a

Answer 1

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