# Question #cb25a

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 ${v}_{\text{avg}}$ 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.

$t = 4.96 \cancel{\min} \cdot \frac{60 s}{1 \cancel{\min}} = 297.6 s$

The distance he walks in time interval 2 is

$s = v \cdot t = 2.02 \frac{m}{s} \cdot 297.6 s = 601.2 m$

To calculate ${v}_{\text{avg}}$ from t = 0 to t = 9.92 min, I will use the total time from both time intervals
${t}_{\text{total}} = 2 \cdot 297.6 s = 595.2 s$

${v}_{\text{avg" = "total distance"/"total time}} = \frac{601.2 m}{595.2 s} = 1.01 \frac{m}{s}$

The question is not clear about the time period for which it wants the ${v}_{\text{avg}}$. So I gave you ${v}_{\text{avg}}$ for time interval 1, time interval 2, and time interval 1+2.

b) His average acceleration ${a}_{\text{avg }}$in the time interval 1 is zero because he is standing still.

I hope this helps,
Steve