# Question #63466

Nov 9, 2017

Displacement is 684 m to the North.

#### Explanation:

I will use v, s, and t from the suvat variable names: v is velocity, s is displacement, and t is time.

Average velocity gives the same result that constant velocity would give, if it were possible to walk and maintain constant velocity when there are flowers to smell and a candy shop's windows to look through.

We can just use the basic form of the velocity equation

${v}_{\text{ave}} = \frac{s}{t}$

So the displacement is given by

$s = {v}_{\text{ave}} \cdot t = 1.2 \frac{m}{s} \cdot 9.5 \min =$

There is a problem there. Time is referred to in seconds for v and in minutes for t. So I will change the units of the time from min to seconds. When that is done, I'll redo that last line with the time units both in seconds.

$9.5 \cancel{\min} \cdot \frac{60 s}{1 \cancel{\min}} = 570 s$

So redoing the line to determine s

$s = {v}_{\text{ave}} \cdot t = 1.2 \frac{m}{\cancel{s}} \cdot 570 \cancel{s} = 684 m$

I hope this helps,
Steve