Question

a horse canters away from a trainer in a straight line, moving 116m 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

at the end of its gallop. the horse is 58 away from its starting point
Therefore, the displacement is = 58 m
and it took a total of 18.8 s to came to its senses
therefore, time taken is = 18.8 s

Average velocity = displacement/time taken
                           = 58 m / 18.8 s
                           = 3/08 m/s

hope this helps

Answer 2

Answer:

The average velocity for the entire trip is 6.30 m/s.

Explanation:

Given that,

Distance $x_{1}= 116\ m$

Time $t_{1}= 14\ s$

A horse canters away from a trainer in a straight line, moving 116 m away in 14 s.

It turns abruptly and gallops halfway back in 4.8 s

(a). We need to calculate the average velocity

Using formula of average velocity

$v_{avg}=\dfrac{total\ displacement}{toatl\ time}$

$v_{avg}=\dfrac{x_{2}-x_{1}}{t_{2}-t_{1}}$

Put the value into the formula

$v_{avg}=\dfrac{116-58}{14-4.8}$

$v_{avg}=6.30\ m/s$

Hence, The average velocity for the entire trip is 6.30 m/s.