Question

a horse canters away from a trainer in a straight line, moving 130m away in 14s. It turns abruptly and gallops halfway back in 4.8s. What's its average velocity for the entire trip and using 'away from trainer' as positive direction?

Answer 1

\[\bar{V_1}=+\frac{130}{14}=+9.286\frac{m}{sec}\]\[\bar{V_2}=-\frac{65}{4.8}=-13.542\frac{m}{sec}\]average velocity...\[\bar{v}=\frac{|\bar{V_1}|+|\bar{V_2}|}{2}=\frac{9.286+13.542}{2}\]\[\bar{v}=11.414\frac{m}{sec}\]