# A horse canters away from its trainer in a straight line, moving 120 m away in 13.0 s. It then...

## Question:

A horse canters away from its trainer in a straight line, moving {eq}120\ m {/eq} away in {eq}13.0\ s {/eq}. It then turns abruptly and gallops halfway back in {eq}4.8\ s {/eq}. Calculate magnitude of the average velocity of the horse for the entire trip.

Using the equation for average velocity, we write:

{eq}v = \frac{d-x}{t_2 - t_1} {/eq}

where:

d is the first position

x is the second position

t is the time

Inserting the values of the available parameters, we write

{eq}v = \frac{(120)-(60)}{(13.0) - (4.8)} \\ v = 7.31 \ m/s {/eq}